Question – Answer Discussion On Electrostatics
A ) Choose the correct answer :
1 ) The force of attraction between two point charges is called as
a ) electrostatic force
b ) magnetic force
c ) gravitational force
d ) nuclear force
Answer : a ) electrostatic force
2 ) 1 C = _________ esu of charge. ( Fill in the blank )
a ) 3 * 1009
b ) 3 * 10-09
c ) 3 * 1090
d ) 3 * 10-90
Answer : a ) 3 * 1009
3 ) 1 emu of charge = ___________ C ( Fill in the blank )
a ) 10
b ) 1 / 10
c ) 1 / ( 3 * 1009 )
d ) ( 1/ 3 ) * 109
Answer : a ) 10
4 ) M-1L-3T4A2 is the dimension of
a ) relative permittivity of a medium
b ) Absolute permittivity of free space
c ) Absolute permittivity of a medium
d ) all
Answer : d ) all
5 ) The force between two charges is 60 N. If the distance between the charges is doubled, what will be the force ?
a ) 15 N
b ) 51 N
c ) 240 N
d ) 420 N
Answer : a ) 15 N
Explanation : From Coulomb’s law of electrostatic force, we have
F = K Q q / r2
keeping K, Q and q constant, we get
F r2 = constant
F r2 = F’ r ‘2
if distance is doubled i.e r’ = 2r
60 * r2 = F’ * ( 2r )2
F’ = 60 / 4 = 15 N
6 ) The force of attraction between two identical charges is 60 N. If the amount of charge is increased to twice of its initial, what will be the force now?
a ) 420 N
b ) 240 N
c ) 204 N
d ) 402 N
Answer :
Explanation : b ) 240 N
From Coulomb’s law of electrostatic of identical point charges,
F = K q2 / r2
Keeping K and r constant
F / q2 = constant
F / q2 = F ‘ / q’2
If the amount of charges is increased to twice of its value i.e q’ = 2q
60 / q2 =F’ / 4q2
F’ = 240 N
7 ) The electric lines of force due to charged particles are
a ) always straight
b ) always curved
c ) sometimes curved
d ) none
Answer : b ) always curved
8 ) The electric field intensity due to a short electric dipole at a point on the axial position is inversely proportional to
a ) square of distance
b ) inverse square of distance
c ) cube of distance
d ) inverse of cube of distance
Answer : d ) inverse of cube of distance
9 ) Torque experienced by an electric dipole placed in an electric field is equal to
a ) dot product of dipole moment and electric field
b ) cross product of dipole moment and electric field
c ) dot product of displacement vector and electric field
d ) cross product of displacement vector and electric field
Answer : b ) cross product of dipole moment and electric field
10 ) when the dipole of dipole moment p is parallel to electric field E, the torque experienced by it, is ________.
a ) 1 unit
b) 0 unit
c ) – 1 unit
d ) pE
Answer : b) 0 unit
11 ) The electric dipole of dipole moment p is placed in an electric field of strength E at an angel of zero degree. What will be the work done due to rotation of the dipole through 90 degree ?
a ) pE
b ) – pE
c ) 0
d ) 1 unit
Answer : a ) pE
12 ) If the dipole makes the angle θ with the electric field and it is rotated through 90 degree, the work done is
a ) dot product of dipole moment and electric field
b ) cross product of dipole moment and electric field
c ) product of dipole moment and electric field
d ) zero
Answer : a ) dot product of dipole moment and electric field
13 ) Which physical quantity is represented by electric lines of force ?
a ) electric charge
b ) electrostatic force
c ) torque experienced by the dipole
d ) electric field intensity
Answer : d ) electric field intensity
14 ) The electric lines of force are
a ) continuous curve
b ) discontinuous curve
c ) straight line
d ) all
Answer : b ) discontinuous curve
15 ) Electric flux means
a ) flow of electric lines of force through a particular surface area perpendicularly
b ) the net number of electric lines of force passing through the specific surface area perpendicularly
c ) dot product of the electric field intensity and the surface area through which the electric lines of force pass
d ) all
Answer : d ) all
16 ) Which physical quantity has V – m as SI unit ?
a ) electric intensity
b ) electric flux
c ) electric force
d ) none
Answer : b ) electric flux
17 ) Electric intensity due to infinite line of charge is
a ) kλ / r
b ) 2Kλ / r
c ) Kr / λ
d ) 2Kr / λ
( where symbols have usual meaning )
Answer : b ) 2Kλ / r
18 ) A field line and an equipotential surface are always
a ) parallel
b ) anti parallel
c ) perpendicular to each other
d ) none
Answer : c ) perpendicular to each other
19 ) The ability of charge bodies to exert force on each other is termed as
a ) electric flux
b ) electric lines of force
c ) electric field intensity
d ) quantisation of charge
Answer : c ) electric field intensity
20 ) Two charges of magnitude -20 C and +0 C are located ( a, 0 ) and ( 4a , 0 ) respectively. What is the electric flux due to these charges through a sphere of radius 3a with its center at origin ?
a ) 2Q / ε ( inwards )
b ) 2Q / ε ( outwards )
c ) ± 2Q / ε
d ) none
( where Q is the net charge enclosed by the surface containing charges )
Answer : a ) 2Q / ε ( inwards )
21 ) Guass theorem is related to
a ) interaction between two point charges
b ) charge distribution
c ) both
d ) none
Answer : b ) charge distribution
22 ) electric potential at a point on the position of an electric dipole moment p is __________
a ) Kq / ( l2 – x2 )
b ) Kp / ( l2 – x2 )
c ) 2Kq / ( l2 – x2 )
d ) 2Kp / ( l2 – x2 )
( where the symbols have usual meanings )
Answer : b ) Kp / ( l2 – x2 )
23 ) The electric potential at any point on the surface of charged spherical conductor is always
a ) zero
b ) undefined
c ) constant
d ) + 1 unit
Answer : c ) constant
24 ) The shape of the equipotential depends on the
a ) quantization of charge
b ) magnitude of charge
c ) distribution of charge
d ) nature of charge
Answer : c ) distribution of charge
25 ) _______________________ perpendicular to the equipotential surface.
a ) electric force is
b ) electric force and electric field are
c ) electric field
d ) none
Answer : b ) electric force and electric field are
26 ) The electric potential at any point on the perpendicular bisector of an electric dipole is always
a ) constant
b ) zero
c ) undefined
d ) +1 unit
Answer : b ) zero
27 ) A point charge of 3μC is located at ( 1 , 0 , 0 ). What is the electric potential at point ( 0 , 1 , 0 ) ?
a ) 27 / √2 kV
b ) 27 √2 kV
c ) 27 / √2 mV
d ) 27 √2 mV
Answer : a ) 27 / √2 kV
Explanation : Electrostatic potential = Kq / r = ( 9 * 109 ) * ( 3 * 10-6 ) / √ ( 12 + 12 + 02 ) = 27 / √2 kV
28 ) If the identical point charges of magnitude +2nC are placed at x = 1, 2 , 4 , 8, 16, ………………… in meter, what will be the electric potential due to these point charges ?
a ) 36 volt
b ) 63 volt
c ) 9 volt
d ) 18 volt
Answer : a ) 36 volt
Explanation :
Electric potential = Kq / r
= ( 9 * 109 ) * ( 2 * 10-9 )( 1 / 1 + 1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 + …………………. ) = 18 { 1 / ( 1 – 1/ 2 ) } = 18 / ( 1 / 2 ) = 36 volt
29 ) Inside the electric conductor, the electric field is zero because
a ) there is no electric lines of force.
b ) net induced positive charges and induced negative charges are balanced by each other.
c ) both
d ) all
Answer : c ) both
30 ) Charge distribution on the surface of the charged conductor may not be same. It depends on the
a ) nature of surface of curvature
b ) direction of the surface area of curvature
c ) nature of charge distribution on the surface of the curvature.
d ) all
Answer : a ) nature of surface of curvature
31 ) The electric charges reside only _______________.
a ) at the surface of conductor
b ) inside the conductor
c ) neither a nor b
d ) both a and b
Ans : a ) at the surface of the conductor
32 ) The strength of electric field at a point on axial line of an electric dipole of very small length is ________ as that at the same distance on its equitorial line
a ) twice
b ) half
c ) thrice
d ) 1 / 3rd
Ans : a ) twice
33 ) The orientation of the electric dipole along the electric field ( p ll E ) corresponds to the ___________.
a ) stable equilibrium
b ) unstable equilibrium
c ) Ionic equilibrium
d ) None
Ans : a ) stable equilibrium
34 ) Electric Field between two infinite plane parallel sheets of charge of surface charge density σ and ( -σ ) :
a ) σ / ε
b ) σ / 2ε
c ) 2σ / ε
d ) 2ε / σ
Ans : a ) σ / ε
35 ) The electric field due to line charge proportional to ___________
a ) r
b ) r2
c ) 1 / r
d ) 1 / r2
Ans : d ) 1 / r2
36 ) If the outward flux through the closed surface is zero, then
a ) there is no charge inside the surface
b ) there may not be equal amounts of positive charges and negative charges inside the surface.
c ) there may be equal amounts of positive and negative charges inside the surface.
d ) none
Ans : a ) there is no charge inside the surface
37 ) When a charge q is placed inside a cube, the total electric flux coming out of the cube is given by
a ) σ / ε
b ) q / ε
c ) σ / ε
d ) 2q / ε
Ans : b ) q / ε
38 ) When a charge q is placed inside a cube, the total electric flux coming out through any one face of the cube is given by ______________.
a ) q / 6ε
b ) 6q / ε
c ) 6σ / ε
d ) σ / 6ε
Ans : a ) q / 6ε
39 ) When a charge q is placed at one of the eight vertices of a cube, the total electric flux coming out through all the six faces of the cube is given by
a ) q / 8ε
b ) 8q / ε
c ) 8qε
d ) 8ε / q
Ans : a ) q / 8ε
40 ) When a charge q ia placed at one of the eight vertices of a cube, the electric flux coming out through any one of the faces is
a ) 24q / ε
b ) q / 24ε
c ) 24qε
d ) 24ε / q
Ans : b ) q / 24ε
B ) Objective Type Questions:
( 1 marks )
1 ) What is the value of absolute permitivity of free space ?
Ans : 8.85 * 10
2 ) Which physical quantity has M-1L-3T4A2 ?
Ans : Absolute permitivity has the given dimension.
3 ) What is the nature of the electric lines of force ?
Ans : The electric lines of force are curve lines or straight lines.
4 ) What is the relation between the electrostatic force and distance between the two point charges ?
Ans : The electrostatic force is inversely proportional to the square of distance between the point charges.
5 ) Give two examples which follow the inverse square law?
Ans : Coulomb’s law of electrostatic force and Newton’s law of gravitation are the two examples which follow the inverse square law.
6 ) Is electrostatic force follows Newton’s third law of motion ?
Ans : Yes, electrostatic force follows Newton’s third law of motion.
7 ) Name the gradient of electric potential.
Ans : The gradient of potential is nothing but electric field intensity.
8 ) What is the value of electric potential at a point 6 cm on the perpendicular bisector of an electric dipole of length 2l ?
Ans : The value of electric potential at a point 6 cm on the perpendicular bisector of an electric dipole of length 2l is zero.
9 ) What will be the electric potential at a point 3 cm away from the center of the spherical conductor of radius 6 cm if +6 µC charge is distributed over the conductor ?
Ans : Required electric potential = K q / r
where K = 9 * 109 SI unit
q = 6 * 10-6 C
r = radius of the conductor = 6 cm = 6 * 10-2 cm
Therefore, Electric potential = { ( 9 * 109 ) * ( 6 * 10-6 ) } / ( 6 * 10-2 ) = 9 * 105 volt.
10 ) Name the locus of all points having same electric potential.
Ans : The name of the locus of all points having same electric potential is equipotential surface.
11 ) What will be the workdone on the equipotential surface having electric field intensity as 0.1 N/C ?
Ans : The work done on the equipotential surface having electric field intensity as 0.1 N / C is zero.
12 ) What will be the velocity of a charged body of charge ‘q’ and of mass ‘m’ in a electric field of electric field of electric potential V ?
Ans : Required velocity = √ ( 2qV / m )
where V = Electric potential.
13 ) What should be the direction of electrostatic field at the surface of a charged conductor ?
Ans : The direction of electrostatic field at the surface of a charged conductor should be perpendicular or normal to the surface.
14 ) Give an example of polar dielectric ?
Ans : Ammonia gas is an example of polar dielectric.
15 ) Is the value of relative permitivity 1 ?
Ans : The value of relative permitivity is always greater than unity or 1.
16 ) What is the CGS unit of capacitance ?
Ans : The CGS unit of capacitance is stat – farad.
17 ) Give two factors influencing the capacitance.
Ans : Proximity of the conductor and surface area of the conductor are the two influencing the capacitance.
18 ) If we want to increase the capacitance which combination should be suitable parallel or series ?
Ans : Parallel combination is suitable to increase the capacitance.
19 ) What is the capacitance of a parallel plate capacitor if the medium is air ?
Ans : Required capacitance = Aε / d
Where A = surface area of each plate
ε = absolute permitvity of free space.
d = distance between the two plates.
20 ) What will be the effect of increment of distance between the parallel plate capacitor on its capacitance ?
Ans : On increment of the distance between the parallel plates of the parallel plate capacitor, its capacitance decreases as it is inversely proportional to the distance.
21 ) How the size of an isolated spherical capacitor affects its capacitance ?
Ans : The capacitance of an isolated spherical capacitor is directly proportional to its size ( radius ) i.e the capacitance increases with increase in size and vice versa.
22 ) 64 small drops of water each of capacitance C and charge q coalesce to form a larger spherical drop. What will be the capacitance of larger spherical drop ?
23 ) What is equipotential surface ?
Ans : It is the locus of all points having same electric potential.
24 ) What is electric flux ?
Ans : The electric flux is defined as the net number of electric lines of force passing through a particular surface area perpendicularly.
25 ) Does electric flux depend on the medium through which the electric lines of force passes ?
Ans : No, electric flux depends on the medium through which the electric lines of force passes.
26 ) What is the value of electric flux through the surface when electric field intensity is normal to the surface area ?
Ans : The value of electric flux through the surface when electric field intensity is normal to the surface area is zero.
27 ) If the value of electric flux is negative, what conclusion one may draw ?
Ans : If the value of electric flux is negative, it indicates that the the electric field is in direction opposite to the surface area.
28 ) Write the mathematical form of Gauss’ theorem
Ans : q / ε is the mathematical form of Gauss’ theorem where q is the amount charge enclosed by the body symmetrically and ε is the permitivity of the medium.
29 ) What is Gaussian surface?
Ans : The Gaussian surface is the surface which encloses the charge symmetrically.
30 ) What is meant by linear charge density ?
Ans : The charge per unit length of the charged body is termed as the linear charge density.
31 ) What is surface charge density ?
Ans : The charge per unit area of the surface of the charged body is called as surface charged density.
32 ) What is volume charge density ?
Ans : The charge per unit volume is named as the volume charge density.
33 ) What does indicate that ” closer the distance between the line of force ” ?
Ans : Closer the distance between the lines of force indicates the higher intensity of the electric filed.
34 ) Are electric lines of force are continuous?
Ans : Electric lines of force are continuous.
35 ) Write down the mathematical expression for the electric torque.
Ans : τ = p x E is the mathematical expression for the electric torque where τ is the torque, p is the electric dipole moment and E is the electric field intensity.
36 ) Which physical quantity is represented by the dot product of electric dipole moment and electric field intensity ?
Ans : Electrostatic work done is represented by the dot product of electric dipole moment and electric field intensity.
37 ) What does q + Q = 0 signify in electrostatic ?
Ans : The given equation implies that q and Q are equal in magnitude but opposite in nature or sign.
38 ) Is electric force is central force in nature ?
Ans : Yes, electric force is central force in nature.
39 ) If qQ > 0, what is the nature of force between the charges ?
Ans : This implies the repulsive nature of force between the charges.
40 ) Can two equipotnetial surface intersect each other ?
Ans : No, two equipotential surface can never intersect each other because at the point of intersection there are two different directions of electric field which is impossible.
41 ) What is test charge ?
Ans : It is vanishingly small positive positive charge that is used to detect the presence of electric field.
42 ) What is the shape of equipotential surfaces for a point charge ?
Ans : The shape of equipotential surfaces for a point charge are concentric spheres whose centers are located at the point charge.
43 ) What is the SI unit of line integral of electric field ?
Ans : Nm / C
44 ) What is electrostatic shielding ?
Ans : The process or phenomenon of protecting a certain region from external electric field is called as Electrostatic shielding.
45 ) In what form is energy stored in a charged capacitor ?
Ans : In a charged capacitor, the energy is stored in the form of potential energy.
46 ) What is the value of dielectric constant for metal ?
Ans : The value of dielectric constant for metal is infinity.
47 ) What is the dipole moment of non – polar molecule ?
Ans : The dipole moment of non – polar molecule is zero.
48 ) If ‘ n ‘ capacitors each of capacitance C have connected in series with a battery of V volt what is the equivalent capacitance ?
Ans : The required equivalent capacitance is n /C.
49 ) When the dielectric slab is introduced between the plates of an isolated parallel plate capacitor, what will be the effect on potential difference ?
Ans : When the dielectric slab is introduced between the plates of an isolated parallel plate capacitor, the potential difference will decrease.
50 ) Two nearby points are at the same potential. What is the intensity of electric field in this region ?
Ans : The intensity of electric field in this region is zero as the potential is constant.
C ) Subjective Type Questions :
1 ) Show that the electric filed intensity on the surface of a charged conductor is E = ( σ / ε )n where σ is the surface density of charge and n is the outward pointing unit normal vector. [ WBCHSE 2018 ]
Ans : From Gauss’ theorem, we have
Φ = q / ε
∮ E . dS = q / ε
Let us assume that the closed surface is homogeneous sphere of radius r.
Let us consider that the electric field intensity is parallel to the surface area, then
E ∮ dS = q / ε
E * ( 4 πr2 ) = q / ε
E = ( q / 4 πr2 ) / ε
E = σ / ε
where σ = q / 4 πr2
Hence, E = ( σ / ε )n
Where n is the outward pointing unit normal vector.
2 ) Why are the electric lines of force not closed loop ? [ WBCHSE 2018 ]
Ans : The electric lines of force originate from the positive charge and end on the negative charge. This lines are mostly straight line or curve. Hence, the the electric lines of force are not closed loop.
3 ) The electric potential at point ( x , y , z ) is given V = -x2y – xz2 + 4 . Find the electric field E at that point. [ WBCHSE 2018 ]
Ans :
let r = xi + yj + zk
r = √ ( x2 + y2 + z2 )
differentiating both sides w.r.t x
dr / dx = 2x / 2√ ( x2 + y2 + z2 )
dr / dx = x / r
dx / dr = r / x
Similarly, dy / dr = r / y and dz / dr = r / z
V = -x2y – xz2 + 4
differentiating both sides with respect to r
dV / dr = -d ( x2y ) / dr – d ( xz2 ) / dr + d ( 4 ) / dr
dV / dr = -2xy( dx/dr ) – x2 ( dy / dr ) – z2 ( dx / dr ) – 2zx ( dz / dr ) + 0
dV / dr = -2xy ( r / x ) – x2 ( r / y ) – z2 ( r / x ) – 2zx ( r / z )
dV / dr = – 2yr – ( x2 r / y ) – ( z2 r / x ) – 2rx
dV / dr = [ -2y – x2 / y – z2 / x – 2x ] r
As we know that dV / dr = E
Therefore, E = [ -2y – x2 / y – z2 / x – 2x ] r
4 ) Show that the electric field intensity is normal to equipotential surface. [ WBCHSE 2018 ]
Ans : As we know that on the equipotential surface, at each and every point the electric potential is same that is the electric potential is constant at each and every point on that surface.
Therefore, V = constant
differentiating both sides with respect to r
dV / dr = d ( constant ) / dr
dV / dr = 0
dV = 0 —————————- ( 1 )
Now, dV =E . dr
E . dr = 0 [ ( by using equation ( 1 ) ]
The dot product of electric field vector and position vector indicates that the angle between them is 90 degree. Hence, the electric field intensity is normal to the surface of the equipotential surface.
5 ) Define surface density of electric charge. Two large conducting spheres carrying charges q and Q are brought close to each other. Is the magnitude of the electrostatic force between them exactly given by qQ / 4 εr2, where r is the distance between the spheres ? [ WBCHSE 2017 ]
Ans : Surface density of electric charge may be defined as the charge per unit surface area of the charged conductor.
No, the magnitude of the electrostatic force between the given conducting spheres carrying charges q and Q when they are brought close to each other as we know from Coulomb’s law of electrostatic, force between the two given spheres = qQ / 4πεr2
6 ) Define dielectric constant. Two charges ± 20 * 10-6 C placed 2 mm apart from an electric dipole. Determine the electric field at at a point 10 cm away from the center of the dipole on its perpendicular bisector. Given : 1 /4πε = 9 *- 109 Nm2C-2 [ WBCHSE 2017 ]
Ans :
Dielectric Constant ( K ) : It is the ratio of the permittivity of the material to the permittivity of the free space. It is also called as relative permittivity or specific inductive constant.
K = ԑ / ԑ0
ԑ is the permittivity of the material and ԑ0 is the permittivity of free space.
Charge ( q ) = ± 20 * 10-6 C
Length of the electric dipole ( 2 l ) = 2 mm = 2 * 10-3 m
Dipole moment ( p ) = q * 2l = ( 20 * 10-6 ) * ( 2 * 10-3 ) = 4 * 10-8 C-m
The required electric field is given by
E = kp / ( r2 + l2 ) 3/2
Where k = 9 * 109 Nm2c-2
r = 10 cm = 10 * 10-2 m
l = ( 2 * 10-3 ) / 2 = 10-3 m
Now, putting all the values we get
E = 359.95 N / C
7 ) Define dielectric polarisation. Deduce an expression for potential energy stored in a parallel plate capacitor. [ WBCHSE 2017 ]
Ans : When an external electric field is applied to an insulating material, the positive charge is aligned in the direction of the electric field and negative charge is aligned against the electric field. Due to this alignment, this insulating material behaves like an electric dipole. This phenomenon is termed as the dielectric polarisation.
Expression for the potential energy stored in a parallel plate capacitor :
Let infinitesimally amount of charge dq is given to a parallel plate capacitor having capacitance C, so that its potential remains constant. Doing so, some work is done. This work is stored in the form of potential energy.
Small work done = dw = Vdq
On integrating both sides, we get
∫ dw = ∫ Vdq
∫dw = ∫ ( q / C ) dq
W = ( 1 / C ) ∫ qdq
W = q2 / 2C + A
Where A is integral constant
If q = 0 then W = 0
Therefore, A = 0
Hence, the potential energy is given by
E = q2 / 2C
Or
E = CV2 / 2
putting C = ԑ A / d, we get
E =ԑ A V2 / 2d
Where A = surface area of the parallel plate capacitor.
V = Potential difference
ԑ = permitivitty of the medium
d = separation of the two parallel plates.
8 ) What is an equipotential surface ? Show that the electric field is always normal to an eauipotential surface.[ WBCHSE 2016 , 2018 ]
Ans : An equipotential surface may be defined as the locus of all the points having same electric potential.
As we know that on the equipotential surface, at each and every point the electric potential is same that is the electric potential is constant at each and every point on that surface.
Therefore, V = constant
differentiating both sides with respect to r
dV / dr = d ( constant ) / dr
dV / dr = 0
dV = 0 —————————- ( 1 )
Now, dV =E . dr
E . dr = 0 [ ( by using equation ( 1 ) ]
The dot product of electric field vector and position vector indicates that the angle between them is 90 degree. Hence, the electric field intensity is normal to the surface of the equipotential surface.
9 ) A glass slab is introduced between the plates of parallel plate capacitor. Does the capacitance of the capacitor increase , decrease or remain uncharged ?
Two capacitors of capacitance 5µF and 10µF are charged to 16 volts and 10 volts respectively. Find the common potential when they are connected in parallel.[ WBCHSE 2016 ]
Ans : A glass slab is introduced between the plates of parallel capacitor, the capacitance of the capacitor increases.
capacitance of first capacitor ( C1 ) = 5µF
potential of first capacitor ( V1 ) = 16 volts
capacitance of second capacitor ( C2 ) = 10µF
potential of second capacitor ( V2 ) = 10 volts
Common potential ( V ) = ( C1V1 + C2V2 ) / ( C1 + C2 ) = {(5 * 10-6 * 16 ) + ( 10 * 10-6 * 10 ) } / { ( 5 + 10 ) * 10-6 } = ( 80 + 100 ) / 15 = 180 / 15 = 12 volts
10 ) 64 identical water drops coalesce to form a large drop. If the nature and amount of the charge be the same for all the drops. Calculate the potential, capacitance and stored energy of the large drop.[ WBCHSE 2016 ]
Ans : No of identical drops ( n ) = 64
Amount of charge on each capacitor = q
Potential of each drop = v
capacitance of each drop ( c ) = 4πԑr
Total charge on the large drop ( Q ) = nq = 64q
We know that,
Volume of n small drops = volume of large drop
n * ( 4π r3 / 3 ) = 4π R3 / 3
64 * r3 = R3
( 4r )3 = R3
R = 4r
Where R is the radius of the larger drop and the r is the radius of each drop.
Now, the capacitance of large ( C ) = 4πԑR = 4πԑ( 4r ) = 4 * ( 4πԑr ) = 4 times of each drop
Potential ( V ) = n2/3 v = ( 43 )2/3v = 16 v = 16 times of the potential of each drop
Stored energy of the larger drop = CV2 / 2 = ( 4c * 16v * 16v ) / 2 = 1024 ( cv2 / 2 ) = 45 energy of each drop.
11 ) Define dielectric dipole moment. Find the torque acting o a dipole when it is place in an uniform electric field E.[ WBCHSE 2015 ]
( here red color indicates that it is vector quantity )
Ans : When an external field is applied to an insulating material, the positive charge is alignment along the direction of the electric field while the negative charge align against the electric field. Due to this, the insulating material behaves like an electric dipole and the dipole moment of this type of dipole is termed as the dielectric dipole moment.
Expression Of The Torque In The Electric Field :
Expression for the Torque experienced by the electric dipole
Torque experienced by an electric dipole placed in an electric dipole is given by
τ = Force * perpendicular distance
τ = q E * x
τ= q E * 2 l sinθ
τ = ( q * 2 l ) * E sinθ
τ = p E sinθ
τ = p × E
where θ is the angle of inclination of the electric dipole, q is the amount of the charges on the ends of the electric dipole and 2 l is the length of the dipole and E is the electric field intensity.
Hence, the torque experienced by an electric dipole placed in an uniform electric field is the cross product of the electric dipole moment and the electric field intensity.
12 ) State Gauss theorem in electrostatics. Find the electric flux through a surface area of 50 m in x – y plane in the electric field E = ( 3i + 2j + k ) V / m.[ WBCHSE 2015 ]
Ans : Gauss theorem states that ” The total electric flux through any closed surface area is equal to 1 / ε0 times the net charge enclosed by the surface. ”
Surface area ( S ) = 50 k
Electric field ( E ) = ( 3i + 2j + k )
Electric flux = E . S = ( 50 k ) . ( ( 3i + 2j + k ) = 50 * 1 = 50 Nm / C
13 ) What is understood by capacitance of a capacitor? A 900 µF capacitor is charged to 100 V by a battery. How much energy is stored in the capacitance ?[ WBCHSE 2015 ]
Ans : The capacitance of a capacitor may be defined as the amount of charge required to raise the potential of a container containing charge. It is the ability of holding or storing amount of charge of a conductor.
Energy stored in the battery = capacitance * ( potential )2 / 2 = ( 900 * 10-6 * 1002 ) / 2 = 4.5 Joule.
Numerical Problems
1 ) Is a charge of 6.5 *10-5 C possible ?
Ans : From quantization of charge, we have
q = ne
where n = no of electron and e is the electronic charge
n = q / e = ( 6.5 *10-5 ) / ( 1.6 * 10-19 ) = 4.0625 * 1014
As this number is whole number, the given charge is possible.
2 ) Prove that the electronic charge is the smallest unit of charge.
Ans : Let us consider that q is the charge smaller than the electronic charge
From quantization of charge we have
n = q / e
⇒ n < 1
which is false as n should be integer.
Hence, there is no charge smaller than the electronic charge.
3 ) Is 0.16 * 10-19 C is possible as charge ?
Ans : n = q / e = ( 0.16 * 10-19 ) / ( 1.6 * 10-19 ) = 0.1
As 0.1 is not an integer, so the given charge is not possible.
4 ) If a body gives out 109 electrons every second, how much time is required to get a total charge of 1 C from it ? [ By S. Chand’s principles of Physics ]
Ans : Total charge ( Q ) = 1 C
Number of electrons in one second ( n ) = 109
Charge flow per second = 1.6 * 10-19 * 109 = 1.6 * 10-10 C
Let the required time is t second
Therefore, Q = net
t = Q / ne = 1 / ( 1.6 * 10-10 ) sec = 6.25 * 109 sec = ( 6.25 * 109 ) / ( 60 * 60 * 24 * 365 ) = 198.19 years
This is the required time.
5 ) What is the electrostatic force between two like point charges of one coulomb each placed 1 m apart in water having dielectric constant as 80 ? What is the nature of the charge ?
Solution :
From Coulomb’s law of electrostatic, the electrostatic force between the two like point charges each of 1 C each placed 1 m apart in water having dielectric constant 80 is given by
F = ( 9 * 109 ) ( 1 * 1 ) / { ( 12 ) * 80 } = 112.5 * 106 N
The nature of the charge is repulsive charge as the two charges are like charges.
6 ) Charges q1 and q2 lie on the x – axis at point x = -4 cm and x = +4 cm respectively. How must q1 and q2 be related so that net electrostatic force on a charge placed at x = + 2 cm is zero.
Solution :
distance between q1 and q2 = 4 – ( -4 ) = 8 cm
distance between q1 and Q = 8 – 2 = 6 cm
distance between q2 and Q = 4 – 2 = 2 cm
At x = +2 cm, the electrostatic force on charge particle Q is zero,
Therefore, Kq1Q / ( 6 * 6 ) = kq2Q / ( 2 * 2 )
( where K is Coulomb’s Constant )
q1 = ( 36 / 4 ) q2
q1 = 9q2
The above equation expresses the the required relationship between the two given charges.
7 ) Charges q and Q are separated by a distance of 8 cm. A third charge is placed at a distance of 3 cm from q. What is the relation between q and Q for equilibrium of charges ?
Solution : Let third charge b x
distance between q and Q = 8 cm
distance between q and x = 3 cm
distance between Q and x = 8 – 3 = 5 cm
At x = +2 cm, the electrostatic force on charge particle Q is zero,
Therefore, Kqx / ( 3 * 3 ) = kQx / ( 5 * 5 )
where K is Coulomb’s Constant
q = ( 25 / 9 ) Q
The above equation expresses the the required relationship between the two given charges.
8 ) Calculate electrostatic force acting on + 2 µC placed at vertex B in the following figure. Angle between AB and BC is 45 degree.
Solution :
AB = 6 cm = 6 * 10-2 m
BC = 7 cm = 7 * 10-2 m
AC = 5 cm = 5 * 10-2 m
Electrostatic force between charges at vertices B and A = FAB = K( 2µ * 2µ ) / AB2 = ( 9 * 109 * 2 * 10-6 * 2 * 10-6 ) / ( 6 * 10-2 )2 = ( 9 * 2 * 2 * 109 -6 -6 ) / ( 36 * 10-2 )= ( 36 * 10-3 ) / ( 36 * 10-4 ) = 10 N
Electrostatic force between charges at vertices B and C = FBC = K( 2µ * 2µ ) / BC2 = ( 9 * 109 * 2 * 10-6 * 2 * 10-6 ) / ( 7 * 10-2 )2 = ( 9 * 2 * 2 * 109 -6 -6 ) / ( 49 * 10-4 )= ( 36 * 10-3 ) / ( 49 * 10-4 ) = 7.35 N
where K is Coulomb’s constant = 9 * 109 SI unit
Angle between FAB and FBC = 1800 – 450 = 1350
Magnitude of resultant force of FAB and FBC = F = ( F2AB + F2BC + 2FABFBC cos135 )1/2 = { 102 + 7.352 + 2 ( 10 ) * ( 7.35 ) cos ( 135 ) }1/2 = { 100 + 54.02 + 147 ( – 0.707 ) }1/2 = ( 154.02 – 103.93 )1/2 = ( 50.09 )1/2 = 7.08 N
9 ) Four equal charges + 6 µC are kept at four vertices of a square of length of each side 20 cm. Calculate the electrostatic force experienced by a charge kept at the center of the square.
Ans :
Let the square is ABCD in which AB = BC = CD = DA = 20 cm = 20 * 10-2 m = 0.2 m
Distance between center O and each corner of the square ( r ) = half of the diagonal = side of the square * √2 / 2 = 0.2√2 / 2 = 0.1 √2 m
Let the charge placed at the center of the square ABCD is q.
Force experienced by q due to charge placed at vertex A = FA = Kq ( 6 * 10-6 ) / AO2 = Kq ( 6 * 10-6 ) / r2
Force experienced by q due to charge placed at vertex A = FB = Kq ( 6 * 10-6 ) / BO2= Kq ( 6 * 10-6 ) / r2
Force experienced by q due to charge placed at vertex A = FC =Kq ( 6 * 10-6 ) / CO2= Kq ( 6 * 10-6 ) / r2
Force experienced by q due to charge placed at vertex A = FD =Kq ( 6 * 10-6 ) / DO2 = Kq ( 6 * 10-6 ) / r2
where K is Coulomb’s constant = 9 * 109 SI unit
Here we see that FA and FC are equal in magnitude but in opposite in direction. So, the effect of these forces cancel each other. Similarly FB and FD also cancel each other.
Therefore, the resultant force acting on the charge placed at the center of the square is zero.
10 ) Three charges q, 3q and -8q are placed at three vertices of an equilateral triangle of length of each side ‘ a ‘. Calculate the electrostatic force acting on -8q.
Solution :
Force experienced by ( -8q ) due to q ( FA ) = K ( 8q ) q / a = 8Kq2 / a2 = 8x
Force experienced by ( -8q ) due to 3q ( FB ) = K ( 8q ) ( 3q ) / a = 24 Kq2 / a2 = 24 x
where x = Kq2 / a2 and K is Coulomb’s constant = 9 * 10 SI unit
Angle between FA and FA = 60 degree.
Resultant force experienced by the charge ( – 8 q ) = ( FA2 + FB2 + 2FAFB cos 60 )1/2 = { ( 8x )2 + ( 24x )2 + 2 ( 8x ) ( 24x ) ( 1 / 2 ) }1/2 = ( 64 x2 + 576 x2 + 192x2 )1/2 = 28.84x = 28.84Kq2 / a2
11 ) A point charge 6 µC is placed in a medium of relative permittivity of 80. What is the electric field intensity at a point 15 cm form the charge ?
Solution : charge ( q ) = 6 µ C = 6 * 10-6 C
Relative permittivity of the medium ( P ) = 80
Distance between the charge and observation point ( r ) = 15 cm = 15 * 10-2 m
Electric field intensity ( E ) = Kq / Pr2 = ( 9 * 10-9 * 6 * 10-6 ) / ( 80 * 15 * 10-2 )2 = 3 * 10-4 N /C
12 ) Electric charges of the same nature and each of magnitude 3 µC are placed at the corners of a square of diagonal 4 m. Find the electrostatic force and electrostatic field at the center of the square.
Solution : At the center of the square, the electrostatic force as well as electrostatic field is zero.
13 ) Four equal charges q are placed at four corners of a rectangle of length l and breadth b. Calculate the electrostatic force experienced by a charge ( – q ) placed at the center of the rectangle.
Solution : Zero
14 ) Two point charges + 9q and + q are separated by a distance of 16 cm. At what point between these charges a third charge Q be placed so that it remains in equilibrium ? ( S. Chand publication )
Solution :
Let the third charge is placed at a distance of x from +9q.
distance between the two given charges ( r ) = 16 cm = 0.16 m
At this point the system is in equilibrium,
K( 9q * Q ) / x2 = K ( Q * q ) / ( r – x )2
( where K is Coulomb’s constant and is equal to 9 * 109 SI unit )
9 / x2 = 1 / ( r – x )2
x2 / 9 = ( r – x )2
x2 / 9 = r2 – 2rx + x2
9r2 – 18rx + 9x2 = x2
9x2 – x2 – 18rx + 9r2 = 0
8x2 – 18 rx + 9r2 = 0
8x2 – ( 12 + 6 )rx + 9r2 = 0
8x2 – 12 rx – 6rx + 9r2 = 0
4x ( 2x – 3r ) – 3r ( 2x – 3r ) = 0
( 2x – 3r ) ( 4x – 3r ) = 0
either
( 2x – 3r ) = 0
2x = 3r
x = 3r / 2
x = ( 3 * 0.16 ) / 2
x = 0.24 m
or ( 4x – 3r ) = 0
4x = 3r
x = 3r / 4
x = 0.12 m
As 0.24 m is out 0.16 m is out of the distance between the two given charges
x ≠ 0.24
Therefore, the required distance is 0.12 m or 12 cm from +9q charge.
15 ) Calculate the force at center o at which 1µC in the following figure if the length of each side of square is 20 cm.
Solution :
Following the figure, we have
FP = Force on charge placed at the center O due to charge placed at P = Kq( 6 µ ) / r2 = 6x
FQ = Force on charge placed at the center O due to charge placed at Q = Kq( 3 µ ) / r2 = 3x
FR = Force on charge placed at the center O due to charge placed at R = Kq( 6 µ ) / r2 = 6x
FS = Force on charge placed at the center O due to charge placed at S = Kq( 3 µ ) / r2 = 3x
Where K is Coulomb’s constant and equal to 9 * 109 SI unit and x = Kq/r2
r = 20√2 / 2 cm = 10√2 cm = 10√2 / 100 m = 0.1√2 m
q = 10-6 C
∴ x = Kqµ / r2 = ( 9 * 109 * 10-6 * 10-6 ) / ( 0.1 √2 )2 = 0.45 SI units
Here, we see that FP and FR as well as FQ and FS ar e parallel to each other.
Resultant Force of FP and FR = F1 = FP + FR = 6x + 6x = 12x
Resultant Force of FS and FQ = F2 = FS + FQ = 3x + 3x = 6x
Here F1 and F2 are acting at an angle of 45 degree.
Resultant force on charge placed center = √ { ( 12x )2 + ( 6x )2 } = √ ( 144x2 + 36x2 ) = x√180 = 0.45 * 13.42 = 6.039 N
16 ) Calculate the electric dipole moment between an electron and a proton 2 Angstrom apart.
Solution : electronic charge ( q ) = 1.6 * 10-19 C
distance between electron and proton ( 2l ) = 2 *10-10 m
electric dipole moment = q * 2l = 1.6 * 10-19 * 2 *10-10 =3.2 *10-29 C – m
17 ) A electric dipole consists of 5µC and -5µC separated by a distance of 10 mm. Calculate electric field intensity ( i ) at point 5 mm from the midpoint of the dipole ( ii ) at point 5 mm from the axial position of the dipole.
Solution : ( i ) the required electric field intensity = E = kp / ( r2 + l2 ) 3/2
where k is Coulomb’s constant and it is equal to 9 * 109 SI unit.
p = dipole moment = 5µC * 10 mm = 5 * 10-6 * 10 *10-3 = 5 * 10-8 C – m
r = distance from the midpoint of the dipole = 5 mm = 5 * 10-3 m
2l = length of the dipole = 10 mm
l = 10 / 2 = 5 mm = 5 * 10-3 m
E = kp / ( r2 + l2 )3/2
E = ( 9 * 109 * 5 * 10-3 ) / { ( 5 * 10-3 )2 + ( 5 * 10-3 )2 }3/2
E = ( 45 * 106 ) / { ( 25 * 10-6 ) + ( 25 * 10-6 ) }3/2
E = ( 45 * 106 ) / ( 50 * 10-6 )3/2
E =