Electrostatic Potential and Capacitance
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A parallel plate capacitor is made up of stair like structure with a plate area A of each stair and that is connected with a wire of length b, as shown in the figure. The capacitance of the arrangement is <!-- [if gte mso 9]><xml>
</xml><![endif]--> the value of x is……………
A parallel plate capacitor is made up of stair like structure with a plate area A of each stair and that is connected with a wire of length b, as shown in the figure. The capacitance of the arrangement is <!-- [if gte mso 9]><xml>
(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
-
Where is a unit vector normal to the surface at a point and s is the surface charge density at that point. (The direction of is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is / .
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
[Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]
(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
-
Where is a unit vector normal to the surface at a point and s is the surface charge density at that point. (The direction of is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is / .
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
[Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]
2.16 Electric field on one side of a charged body is and electric field on the other side of the same body is If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,
= ………….(i)
Where,
=
2.31 (a) Two large conducting spheres carrying charges and are brought close to each other. Is the magnitude of electrostatic force between them exactly given by /4 , where r is the distance between their centres?
(b) If Coulomb’s law involved 1/ dependence (instead of 1/ ), would Gauss’s law be still true ?
(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?
(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?
(f) What meaning would you give to the capacitance of a single conductor?
(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).
2.31 (a) Two large conducting spheres carrying charges and are brought close to each other. Is the magnitude of electrostatic force between them exactly given by /4 , where r is the distance between their centres?
(b) If Coulomb’s law involved 1/ dependence (instead of 1/ ), would Gauss’s law be still true ?
(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?
(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?
(f) What meaning would you give to the capacitance of a single conductor?
(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).
2.31 The force between two conducting spheres is not exactly given by the expression /4 , because there is non-uniform charge distribution on the spheres.
Gauss's law will not be true, if Coulomb's law involved dependence, instead of , on r
Yes. If a small t
A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations.
A. Key K is kept closed and plates of capacitors are moved apart using insulating handle.
B. Key K is opened and plates of capacitors are moved apart using insulating handle.
Choose the correct option(s).
(a) In A, Q remains the same but G changes
(b) In B, V remains the same but C changes
(c) In A, V remains the same hence Q changes
(d) In B ,Q remains the same hence V changes
A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations.
A. Key K is kept closed and plates of capacitors are moved apart using insulating handle.
B. Key K is opened and plates of capacitors are moved apart using insulating handle.
Choose the correct option(s).
(a) In A, Q remains the same but G changes
(b) In B, V remains the same but C changes
(c) In A, V remains the same hence Q changes
(d) In B ,Q remains the same hence V changes
This is a multiple choice answer as classified in NCERT Exemplar
(c), (d) Case A: When key K is kept closed and plates of capacitors are moved apart using insulating handle.
The battery maintains the potential difference across connected capacitor in every circumstance. The separation between tw
If a conductor has a potential V≠0 and there are no charges anywhere else outside, then
(a) There must be charges on the surface or inside itself
(b) There cannot be any charge in the body of the conductor
(c) There must be charges only on the surface
(d) There must be charges inside the surface
If a conductor has a potential V≠0 and there are no charges anywhere else outside, then
(a) There must be charges on the surface or inside itself
(b) There cannot be any charge in the body of the conductor
(c) There must be charges only on the surface
(d) There must be charges inside the surface
This is a multiple choice answer as classified in NCERT Exemplar
(a, b) The potential of a body is due to charge of the body and due to the charge of surrounding. If there are no charges anywhere else outside, then the potential of the body will be due to its own charge. If there is a cavity inside
In the circuit shown in figure initially key K1is closed and key K2 is open. Then K1 is opened and K2 is closed (order is important). [Take Q’1 and Q’2 as charges on C1 and C2 and V1 and V2 as voltage respectively.] Then, E
(a) Charge on C, gets redistributed such that V1 = V2
(b) Charge on C1 gets redistributed such that Q’1 = Q’2
(c) Charge on C1 gets redistributed such that C1V1 + C2V2 = C1E
(d) Charge on C1 gets redistributed such that Q’1 + Q’2=Q

In the circuit shown in figure initially key K1is closed and key K2 is open. Then K1 is opened and K2 is closed (order is important). [Take Q’1 and Q’2 as charges on C1 and C2 and V1 and V2 as voltage respectively.] Then, E
(a) Charge on C, gets redistributed such that V1 = V2
(b) Charge on C1 gets redistributed such that Q’1 = Q’2
(c) Charge on C1 gets redistributed such that C1V1 + C2V2 = C1E
(d) Charge on C1 gets redistributed such that Q’1 + Q’2=Q

This is a multiple choice answer as classified in NCERT Exemplar
(a), (d) Initially key K1 is closed and key K2 is open, the capacitor C1 is charged by battery and capacitor C2 is still uncharged. Now K1 is opened and K2 is closed, the capacitors C1 and C2 both are
In a region of constant potential
(a) The electric field is uniform
(b) The electric field is zero
(c) There can be no charge inside the region
(d) The electric field shall necessarily change if a charge is placed outside the region
In a region of constant potential
(a) The electric field is uniform
(b) The electric field is zero
(c) There can be no charge inside the region
(d) The electric field shall necessarily change if a charge is placed outside the region
This is a multiple choice answer as classified in NCERT Exemplar
(b), (c) We know, the electric field intensity E and electric potential V are dV related as E =- dV/dr or we can write |E|=ΔV/Δr
The electric field intensity E and electric potential V are related as E = 0 and for V = constant
The work done to move a charge along an equipotential from A to B
The work done to move a charge along an equipotential from A to B
This is a multiple choice answer as classified in NCERT Exemplar
(b), (c) If potential of two point is same then work done is also zero according to relation As W=qdv=q (final potential-initial potential).
Also W=-qdv= -q
For equipotential surface V2-V2=0 and W=0
Equipotential surfaces
(a) are closer in regions of large electric fields compared to regions of lower electric fields
(b) will be more crowded near sharp edges of a conductor
(c) will be more crowded near regions of large charge densities
(d) will always be equally spaced
Equipotential surfaces
(a) are closer in regions of large electric fields compared to regions of lower electric fields
(b) will be more crowded near sharp edges of a conductor
(c) will be more crowded near regions of large charge densities
(d) will always be equally spaced
This is a multiple choice answer as classified in NCERT Exemplar
(a), (b), (c) Properties of equipotential surfaces
Consider a uniform electric field in the z -direction. The potential is a constant
(a) in all space
(b) for any x for a given z
(c) for any y for a given z
(d) on the x-y plane for a given z

Consider a uniform electric field in the z -direction. The potential is a constant
(a) in all space
(b) for any x for a given z
(c) for any y for a given z
(d) on the x-y plane for a given z

This is a multiple choice answer as classified in NCERT Exemplar
(c) Field lines are always perpendicular to the direction of equipotential surface. The electric field in the in z- direction suggest that equipotential surface are in x-y plane. The shape of equipotential surface depends upon nature a
