Electrostatic Potential and Capacitance Notes

Introduction to Electrostatic Potential

Potential energy was introduced (Chapters 5 and 7, Class XI).
Work done by an external force against a conservative force (spring or gravity) is stored as potential energy.
When the external force is removed, potential energy converts to kinetic energy, conserving the total energy.
Examples of conservative forces: spring force, gravitational force, and Coulomb force.
Coulomb force is conservative due to its inverse-square dependence on distance, similar to gravitational force.

Electrostatic Potential Energy

Analogous to gravitational potential energy, electrostatic potential energy can be defined for a charge in an electrostatic field.
Consider an electrostatic field E due to a charge configuration.
For simplicity, consider E due to a charge Q at the origin.
Imagine bringing a test charge q from point R to point P against the repulsive force from Q (assuming both are positive).
Assume q is small enough not to disturb the original configuration (or Q is held fixed).
Apply an external force F{ext} to counter the electric force FE (F{ext} = -FE).
The charge q is brought from R to P with infinitesimally slow constant speed (no net force or acceleration).
Work done by F{ext} is the negative of work done by FE and is stored as potential energy of q.
If F{ext} is removed at P, FE will move q away from Q, converting potential energy to kinetic energy.

Work Done and Potential Energy Difference

Work done by external forces in moving a charge q from R to P is described by the equation: W{RP} = -\int{R}^{P} F_E \cdot dl
W_{RP} is against the electrostatic repulsive force and is stored as potential energy.
A charge q in an electric field possesses electrostatic potential energy.
The potential energy difference between points R and P is: \Delta U = UP - UR = W_{RP}
Displacement is opposite to the electric force, so work done by the electric field is negative ($-W_{RP}).
Electric potential energy difference is the work done by an external force in moving charge q from one point to another without acceleration.

Comments on Potential Energy

(i) The right side of the equation depends only on the initial and final positions of the charge; the electrostatic field's work depends only on the initial and final points, independent of the path.
This path-independence is a fundamental characteristic of conservative forces.
Potential energy is meaningful only if work is path-independent.
Path-independence can be proved using Coulomb’s law (proof omitted).
(ii) Potential energy difference is defined in terms of work, a physically meaningful quantity.
Potential energy is undetermined to within an additive constant, meaning the actual value is not physically significant, only the difference is.
Adding a constant \alpha to potential energy at every point does not change the potential energy difference: (UP + \alpha) - (UR + \alpha) = UP - UR
There is freedom in choosing the point where potential energy is zero.
A convenient choice is to have electrostatic potential energy zero at infinity.
With this choice, setting point R at infinity, we get: UP = W{\infty P}
Potential energy of charge q at a point is the work done by the external force in bringing q from infinity to that point.

Electrostatic Potential

Electrostatic potential is characteristic of the electric field.
Work done per unit test charge is independent of q.
From Eq. (2.1): VP - VR = \frac{UP - UR}{q}
VP and VR are electrostatic potentials at P and R, respectively.
Only potential difference is physically significant.
Choosing potential to be zero at infinity:
Work done by an external force in bringing a unit positive charge from infinity to a point = electrostatic potential (V) at that point.
Mathematically, V = \frac{W_{\infty P}}{q}
To obtain work done per unit test charge, use an infinitesimal test charge \delta q, obtain work done \delta W in bringing it from infinity, and determine the ratio \frac{\delta W}{\delta q}
The external force at every point must be equal and opposite to the electrostatic force.

Potential Due to a Point Charge

Consider a point charge Q at the origin (take Q > 0).
Determine the potential at point P with position vector r from the origin.
Calculate the work done in bringing a unit positive test charge from infinity to point P.
For Q > 0, work done against the repulsive force on the test charge is positive.
Work done is independent of the path; choose a convenient radial path from infinity to P.
At an intermediate point P' on the path, the electrostatic force on a unit positive charge is:
\frac{1}{4 \pi \epsilon_0} \frac{Q}{r'^2} \hat{r}'
Work done against this force from r' to r' + \Delta r' is:
\Delta W = - \frac{Q}{4 \pi \epsilon_0} \frac{\Delta r'}{r'^2}
The negative sign appears because for \Delta r' < 0, \Delta W is positive.
Total work done (W) by the external force is obtained by integrating:
W = - \int{\infty}^{r} \frac{Q}{4 \pi \epsilon0} \frac{dr'}{r'^2} = \frac{Q}{4 \pi \epsilon_0 r}
By definition, this is the potential at P due to the charge Q:
V(r) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}

Sign of Charge and Potential

Equation is true for any sign of Q, though we considered Q > 0 in its derivation.
For Q < 0, V < 0, i.e., work done (by the external force) per unit positive test charge is negative.
The work done by the electrostatic force in bringing the unit positive charge from infinity to P is positive.
Since for Q < 0, the force on a unit positive test charge is attractive, the electrostatic force and the displacement (from infinity to P) are in the same direction.
Equation is consistent with the choice that potential at infinity be zero.

Variation of Potential and Field with Distance

Electrostatic potential (V \propto \frac{1}{r}) varies inversely with r.
Electrostatic field (E \propto \frac{1}{r^2}) varies inversely with r^2.

Example 2.1

(a) Calculate the potential at a point P due to a charge of 4 \times 10^{-7} C located 9 cm away.
(b) Obtain the work done in bringing a charge of 2 \times 10^{-9} C from infinity to point P.
Does the answer depend on the path?

Solution:

(a) V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r} = (9 \times 10^9 Nm^2C^{-2}) \frac{4 \times 10^{-7} C}{0.09 m} = 4 \times 10^4 V
(b) W = qV = (2 \times 10^{-9} C)(4 \times 10^4 V) = 8 \times 10^{-5} J
No, work done is path independent.

The Electric Dipole

An electric dipole consists of two charges q and -q separated by a small distance 2a.
Its total charge is zero.
A dipole is characterized by a dipole moment vector p whose magnitude is q \times 2a and which points from -q to q.
The electric field depends on the angle between r and p.
The field falls off at a large distance, not as 1/r^2 (typical of single charge), but as 1/r^3.

Electric Potential Due to a Dipole

Electric potential follows the superposition principle.
The potential due to the dipole is the sum of potentials due to charges q and -q:
V = \frac{1}{4 \pi \epsilon0} \left( \frac{q}{r1} - \frac{q}{r_2} \right)
Where r1 and r2 are the distances of the point P from q and -q respectively.
By geometry,
r1^2 = r^2 + a^2 - 2ar \cos \theta r2^2 = r^2 + a^2 + 2ar \cos \theta
Assuming r \gg a,
r1 \cong r - a \cos \theta r2 \cong r + a \cos \theta
Using the Binomial theorem, and retaining terms up to the first order in a/r, we obtain
\frac{1}{r1} \cong \frac{1}{r} \left( 1 + \frac{a}{r} \cos \theta \right) \frac{1}{r2} \cong \frac{1}{r} \left( 1 - \frac{a}{r} \cos \theta \right)
So, V = \frac{q}{4 \pi \epsilon_0} \frac{2a \cos \theta}{r^2}
Since p = 2qa, V = \frac{1}{4 \pi \epsilon_0} \frac{p \cos \theta}{r^2}
Since p \cos \theta = p.\hat{r}, electric potential of a dipole is given by
V = \frac{1}{4 \pi \epsilon_0} \frac{p.\hat{r}}{r^2} , (r \gg a).
Potential on the dipole axis (\theta = 0, \pi) is:
V = \pm \frac{1}{4 \pi \epsilon_0} \frac{p}{r^2}
(Positive sign for \theta = 0, negative sign for \theta = \pi).
The potential in the equatorial plane (\theta = \pi/2) is zero.

Contrasting Features of Dipole Potential

(i) The potential depends not just on r but also on the angle between the position vector r and the dipole moment vector p.
(ii) The electric dipole potential falls off, at a large distance, as 1/r^2, not as 1/r, characteristic of the potential due to a single charge.

Potential Due to a System of Charges

Consider a system of charges q1, q2, …, qn with position vectors r1, r2, …, rn.
The potential V1 at P due to charge q1 is:
V1 = \frac{1}{4 \pi \epsilon0} \frac{q1}{r{1P}}
r{1P} is the distance between q1 and P.
By the superposition principle, the potential V at P is the algebraic sum of the potentials:
V = V1 + V2 + … + V_n
V = \frac{1}{4 \pi \epsilon0} \left( \frac{q1}{r{1P}} + \frac{q2}{r{2P}} + … + \frac{qn}{r_{nP}} \right)
For a continuous charge distribution characterized by a charge density \rho(r), we have:
V = \frac{1}{4 \pi \epsilon_0} \int \frac{\rho(r')}{r - r'} dv'

Uniformly Charged Spherical Shell

Electric field outside the shell is as if the entire charge is concentrated at the center, so potential outside is:
V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r}, (r \ge R)
q is the total charge and R is its radius.
Electric field inside the shell is zero.
Potential is constant inside the shell and equals its surface value:
V = \frac{1}{4 \pi \epsilon_0} \frac{q}{R}

Example 2.2

Two charges 3 \times 10^{-8} C and -2 \times 10^{-8} C are located 15 cm apart.
Find the point on the line joining the two charges where the electric potential is zero.

Solution:

Let the origin O be at the positive charge.
Let P be the required point at x.
V = \frac{1}{4 \pi \epsilon_0} \left( \frac{3 \times 10^{-8}}{x} - \frac{2 \times 10^{-8}}{15 - x} \right) = 0
x = 45 cm

Example 2.3

Figures 2.8 (a) and (b) show the field lines of a positive and negative point charge respectively.
- (a) Signs of potential differences VP – VQ, VB – VA?
- (b) Sign of potential energy difference of a negative charge between Q and P; A and B?
- (c) Sign of work done by the field moving a positive charge from Q to P?
- (d) Sign of work done by external agency moving a negative charge from B to A?
- (e) Does a negative charge's kinetic energy increase or decrease going from B to A?

Solution:

(a) VP – VQ is positive. VB – VA is Positive.
(b) Potential energy difference between Q and P is positive and between A and B is positive.
(c) Work done by the field is negative.
(d) Work done by external agency is positive.
(e) Kinetic energy decreases.

Equipotential Surfaces

An equipotential surface is a surface with a constant value of potential at all points.
For a single charge q, equipotential surfaces are concentric spherical surfaces.
Electric field lines are radial; the electric field is normal to the equipotential surface.
In general, the equipotential surface is normal to the electric field.
If the field were not normal, a non-zero component would exist along the surface, requiring work to move a test charge.
For a uniform electric field E along the x-axis, equipotential surfaces are planes normal to the x-axis.

Relation Between Field and Potential

Consider two closely spaced equipotential surfaces A and B, with potentials V and V + \Delta V.
\Delta l is the perpendicular distance from B to A.
Moving a unit positive charge from B to A against the field, work done is |E| \Delta l.
|E| \Delta l = V – (V + \Delta V) = -\Delta V
|E| = - \frac{\Delta V}{\Delta l}
Electric field is in the direction of the steepest potential decrease.
Its magnitude is the change in potential per unit displacement normal to the equipotential surface.

Potential Energy of a System of Charges

Consider two charges q1 and q2 with position vectors r1 and r2.
Work done to bring q1 from infinity to r1 is zero.
The work to bring q2 from infinity to r2 is q2 times the potential at r2 due to q1. \frac{1}{4 \pi \epsilon0} \frac{q1 q2}{r_{12}}
r_{12} is the distance between points 1 and 2.
Potential energy of the system is U = \frac{1}{4 \pi \epsilon0} \frac{q1 q2}{r{12}}.
If q1 q2 > 0, potential energy is positive (repulsive force).
If q1 q2 < 0, potential energy is negative (attractive force).

Generalization for Three Charges

For three charges q1, q2 and q3 located at r1, r2, r3:U = \frac{1}{4 \pi \epsilon0} \left( \frac{q1 q2}{r{12}} + \frac{q1 q3}{r{13}} + \frac{q2 q3}{r{23}} \right)

Example 2.4

Four charges at the corners of a square ABCD of side d.
- (a) Work required to put together the arrangement.
- (b) Extra work needed to bring q_0 to the center E.

Solution:

(a) Work = \frac{q^2}{4 \pi \epsilon_0} \left( \frac{-4}{d} + \frac{2}{\sqrt{2}d} \right)
(b) The potential at center is zero. So, no extra work is needed.

Potential Energy in an External Field

What is the potential energy of a charge q in a given field?
The external field E is not produced by the given charge(s).
The external electric field E and corresponding external potential V may vary from point to point.
V at a point P is the work done in bringing a unit positive charge from infinity to P.
Work done in bringing a charge q from infinity to P is qV.
Potential energy of q at r in an external field = qV(r).
If an electron with charge q = e = 1.6 \times 10^{-19} C is accelerated by a potential difference of \Delta V = 1 volt, it gains energy of q \Delta V = 1.6 \times 10^{-19}J.
This unit of energy is defined as 1 electron volt or 1eV, i.e., 1 eV = 1.6 \times 10^{-19}J.

Potential Energy of a System of Two Charges

Potential energy of a system of two charges q1 and q2 located at r1 and r2 respectively, is:
U = q1 V(r1) + q2 V(r2) + \frac{1}{4 \pi \epsilon0} \frac{q1 q2}{r{12}}
r{12} is the distance between q1 and q_2.

Example 2.5

(a) Electrostatic potential energy of a system of two charges 7 \mu C and -2 \mu C at (–9 cm, 0, 0) and (9 cm, 0, 0).
(b) Work required to separate them infinitely far apart.
(c) Electrostatic energy of the same in an external field E = A (1/r^2), A = 9 \times 10^5 NC^{-1} m^2

Solution:

(a) -0.7 J
(b) 0.7J
(c) The net electrostatic energy is 49.3 J

Potential Energy of a Dipole in an External Field

Consider a dipole with charges q1 = +q and q2 = –q in a uniform electric field E.
It experiences no net force, but experiences a torque:
\tau = p \times E
The torque rotates it (unless p is parallel or antiparallel to E).

Potential Energy of the Dipole

External torque \tau{ext} rotates the dipole from angle \theta0 to angle \theta:
W = \int^{\theta}{\theta0} \tau{ext} d \theta = pE (\cos \theta0 - \cos \theta)
With \theta_0 = \pi/2:
U(\theta) = -pE \cos \theta = -p.E

Eq. (2.29) to Dipole

U = q [V(r1) – V (r2)] + \frac{1}{4 \pi \epsilon_0} \frac{q}{2a \cos \theta}
$[V(r1)–V (r2)] = –E \times 2a \cos \theta .
Then U = -p.E

Example 2.6

Molecule has permanent dipole moment 10^{-29} Cm. A mole polarized by a strong field of 10^6 Vm^{-1}. Field is suddenly changed by 60^\circ. Heat released?

Solution:

Heat released = 3J

Electrostatics of Conductors

Conductors contain mobile charge carriers.
In metallic conductors, charge carriers are electrons.

Important Results Regarding Electrostatics of Conductors

1. Inside a conductor, the electrostatic field is zero.
2. At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point.
3. The interior of a conductor can have no excess charge in the static situation.
4. Electrostatic potential is constant throughout the volume of the conductor and has the same value on its surface.
5. Electric field at the surface of a charged conductor is \vec{E} = \frac{\sigma}{\epsilon_0} \hat{n}
where \sigma is the surface charge density and \hat{n} is a unit vector normal to the surface in the outward direction.
6. Electrostatic shielding

Example 2.7

(a) A comb rubbed through dry hair attracts bits of paper. Why?
(b) Ordinary rubber is an insulator. Special tyres are slightly conductive. Why?
(c) Vehicles carrying flammable materials have ropes touching the ground. Why?
(d) A bird perches on a high power line, and nothing happens. A man touching the line gets a shock. Why?

Solution:

(a)Comb is charged and polarizes molecules in paper.
(b)Conduct charge to the ground.
(c)Similar to (b).
(d)Potential difference.

Dielectrics and Polarization

Dielectrics are non-conducting substances.

Polar and Non-Polar Molecules

In a non-polar molecule, the centers of positive and negative charges coincide.
Examples are Oxygen (O2) and Hydrogen (H2).
A polar molecule has separated centers of positive and negative charges.
Examples are HCl or Water (H_2O).

Effect of External Electric Field

In an external electric field, non-polar molecules develop an induced dipole moment.
Polar molecules align with the field.
The dipole moment per unit volume is called Polarization ( P ).
For linear isotropic dielectrics, P = \epsilon0 \chie E

Surface Charges

At the surfaces of the dielectric normal to the electric field, there is a net charge density (\pm \sigma_p ).
*(arises from bound charges)
The total field is thereby reduced.

Capacitors and Capacitance

A capacitor is a system of two conductors separated by an insulator.
The relation between charge and potential is: Q = CV.
C is called the capacitance of the capacitor.
The SI unit of capacitance is 1 farad (=1 coulomb volt-1) or 1 F = 1 C V–1.
The maximum electric field that a dielectric medium can withstand without break-down (of its insulating property) is called its dielectric strength; for air it is about 3 \times 10^6 Vm^{-1}.

Parallel Plate Capacitor

Parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance.
Using Eq. (1.33), the electric field in different regions is:
Outer region I. E = 0
Outer region II. E = 0
In the inner region between the plates 1 and 2, the electric fields due to the two charged plates add up, giving
E = \frac{\sigma}{\epsilon_0}.
The direction of the electric field is from the positive to the negative plate.
Now for a uniform electric field, the potential difference is:
V = Ed = \frac{1}{\epsilon_0} \frac{Qd}{A}
The capacitance C of the parallel plate capacitor is then:
C = \frac{Q}{V}
C = \frac{\epsilon_0 A}{d}

Effect of Dielectric on Capacitance

Suppose that the medium between the plates is fully occupied by the intervening region.
The electric field in the dielectric then corresponds to the case when the net surface charge density on the plates is ±( \sigma - \sigma_P ).
Dielectric constant is K =\frac{ \sigma }{ \sigma - \sigma_P }
With dielectric:
V = \frac{Qd}{KA \epsilon0 } C = \frac{K \epsilon0 A}{d}
\epsilon = K \epsilon_0 (permittivity of the medium
K (dielectric constant)

Example 2.8

A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor.
It has thickness (3/4)d, where d is the separation of the plates
How is the capacitance changed when the slab is inserted between the plates?

Solution:

Capacitance increases C = \frac{4C_0 K}{K+3}

Combination of Capacitors - Series

Charges on all plates are same. Q
V = V1 + V2, with V1 = Q/C1 and V2 = Q/C2
C = \frac{(C1 C2 )}{C1 + C2}
In general, 1/C = 1/C1 + 1/C2 + .. + 1/Cn

Combination of Capacitors - Parallel

The potential difference is the same: V
The total charge is now Q = Q1 + Q2
Hence C = C1 + C2
In general, C = C1 + C2 + … + Cn

Example 2.9

A network of four 10 \mu F capacitors is connected to a 500 V supplyDetermine (a) the equivalent capacitance of the network and (b) the charge on each capacitor

Answer

(a) The equivalent capacitance is C =13.3\mu F. (b)The charge on each of the capacitors,C1, C2 and C3 is Q = 1.7 \times 10^{-3}C, and Q’ = 5.0 \times 10^{-3} C

Energy Stored In a Capacitor

Consider two uncharged conductors 1 and 2. Imagine next a process of transferring charge from conductor 2 to conductor 1 bit by bit, so that at the end, conductor 1 gets charge Q. By charge conservation, conductor 2 has charge –Q at the end .
Work is stored in potential energy U = \frac{1}{2} \frac{Q^2}{C} = \frac{1}{2} CV^2 = \frac{1}{2} QV$$
Since electrostatic force is conservative, this work is stored in the form of potential energy of the system. For the same reason, the final result for potential energy
*
Energy density u =(1/2)0E2

Example 2.10

(a) A 900 pF capacitor is charged by 100 V battery. How much electrostatic energy is stored by the capacitor? (b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor. What is the electrostatic energy stored by the system?

Answer

(a) The charge on the capacitor is Q = CV = 9 \times 10-8 C , The energy stored by the capacitor 4.5 \times 10-6 J
(b) The total energy of the system 2.25 \times 10-6 J. In going from (a) to (b), though no charge is lost; the final energy is only half the initial energy. There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in the form of heat and electromagnetic radiation.