Electrostatic Potential and Capacitors

Introduction to Electrostatic Potential

Potential energy was introduced in Chapters 5 and 7 of Class XI.
Work done by an external force in moving a body against a conservative force (like spring or gravity) is stored as potential energy.
When the external force is removed, the body gains kinetic energy while losing potential energy, conserving the total energy.
Conservative forces: Spring force, gravitational force, and Coulomb force.
Coulomb force, like gravitational force, has an inverse-square dependence on distance.
Electrostatic potential energy can be defined for a charge in an electrostatic field, similar to gravitational potential energy.
Consider an electrostatic field E due to a charge configuration.
Field E due to a charge Q at the origin.
A test charge q is brought from point R to point P against the repulsive force from charge Q.
Assume q is small enough not to disturb the original configuration (charge Q at the origin). If not, Q is fixed at the origin by an external force.
External force F{\text{ext}} is applied to counter the repulsive electric force FE (F{\text{ext}} = -FE).
Charge q is brought from R to P at an infinitesimally slow constant speed, meaning no net force or acceleration acts on it.
Work done by the external force is the negative of the work done by the electric force and is stored as potential energy of charge q.
If the external force is removed at P, the electric force moves the charge away from Q, converting potential energy into kinetic energy while conserving total energy.
Work done by external forces in moving charge q from R to P: W{RP} = -\intR^P F_E \cdot dl
This work is against the electrostatic repulsive force and is stored as potential energy.
A charge q in an electric field possesses electrostatic potential energy, and the work done increases its potential energy by the potential energy difference between points R and P.
Potential energy difference: \Delta U = UP - UR = W_{RP}. Note that the displacement is opposite to the electric force, hence the negative sign for the work done by the electric field.
Electric potential energy difference between two points is defined as the work required by an external force to move charge q from one point to another without acceleration.
The work done depends only on the initial and final positions of the charge, not on the path taken.
Path-independence is a characteristic of a conservative force.
Potential energy would not be meaningful if work depended on the path.
Path-independence of work done by an electrostatic field can be proven using Coulomb’s law.

Potential Energy and Zero Potential

Potential energy difference is defined in terms of work, a physically meaningful quantity.
Potential energy is undetermined to within an additive constant; only the difference in potential energy is physically significant.
Adding an arbitrary 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 set electrostatic potential energy to zero at infinity.
With this choice, if point R is at infinity:
W{\infty P} = UP - U{\infty} = UP (since U_{\infty} = 0)
Potential energy of charge q at a point is the work done by an external force in bringing the charge from infinity to that point.

Electrostatic Potential

For a general static charge configuration, potential energy of a test charge q is defined in terms of work done on q.
The work is proportional to q, since the force at any point is qE (E is the electric field due to the charge configuration).
Work per unit test charge is characteristic of the electric field.
Electrostatic potential V is the work done per unit test charge.
Work done by external force in bringing a unit positive charge from point R to P: VP - VR = \frac{UP - UR}{q}.
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.
With an infinitesimal test charge \Delta q, work done \Delta W in bringing it from infinity to a point, the ratio \frac{\Delta W}{\Delta q} is determined.
The external force at every point of the path is equal and opposite to the electrostatic force on the test charge.

Potential Due to a Point Charge

Consider a point charge Q at the origin. Assume Q > 0.
Determine the potential at any 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 is positive.
Workdone is independent of the path so take a radial path from infinity to P.
At 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 is obtained by integrating from r' = \infty to r' = r:
W = - \int{\infty}^{r} \frac{Q}{4 \pi \epsilon0} \frac{dr'}{r'^2} = \frac{Q}{4 \pi \epsilon0} \left[ \frac{1}{r'} \right]{\infty}^{r} = \frac{Q}{4 \pi \epsilon_0 r}
Potential at P due to charge Q:
V(r) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}
Equation is true for any sign of Q.
For Q < 0, V < 0: work done by the external force is negative.
Work done by the electrostatic force is positive (attractive force).
Equation is consistent with the choice that potential at infinity is zero.
Electrostatic potential ( \propto \frac{1}{r} ) and electrostatic field ( \propto \frac{1}{r^2} ) vary with r.

Example 2.1

Calculate the potential at a point P due to a charge of 4 \times 10^{-7} C located 9 cm away. (b) Hence obtain the work done in bringing a charge of 2 \times 10^{-9} C from infinity to the point P. Does the answer depend on the path along which the charge is brought?
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
W = qV = 2 \times 10^{-9}C \times 4 \times 10^4V = 8 \times 10^{-5} J
Work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along r and another perpendicular to r. The work done corresponding to the later will be zero.

Potential Due to an Electric Dipole

Electric dipole: two charges q and -q separated by a small distance 2a.
Total charge is zero.
Dipole moment vector p: magnitude is q \times 2a, direction from -q to q.
Electric field of a dipole at position vector r depends on magnitude r and the angle between r and p.
Electric field falls off as \frac{1}{r^3} at large distances.
Determine the electric potential due to a dipole and contrast with potential due to a single charge.
Origin at the center of the dipole.
Electric field obeys the superposition principle, so electrostatic potential also follows superposition.
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)
r1 and r2 distances of point P from q and -q respectively.
r_1^2 = r^2 + a^2 - 2ar \cos \theta
r_2^2 = r^2 + a^2 + 2ar \cos \theta
For r >> a, retain terms up to the first order in \frac{a}{r}:
r1 \cong r - a \cos \theta r2 \cong r + a \cos \theta
\frac{1}{r_1} = \frac{1}{r} (1 - \frac{a}{r} \cos\theta )^{-1} \approx \frac{1}{r} (1 + \frac{a}{r} \cos\theta )
\frac{1}{r_2} = \frac{1}{r} (1 + \frac{a}{r} \cos\theta )^{-1} \approx \frac{1}{r} (1 - \frac{a}{r} \cos\theta )
Substituting and using binomial theorem:
V = \frac{q}{4 \pi \epsilon0} \left[ \frac{1}{r} (1 + \frac{a}{r} \cos\theta ) - \frac{1}{r} (1 - \frac{a}{r} \cos\theta ) \right] V = \frac{q}{4 \pi \epsilon0} \left[ \frac{2a}{r^2} \cos\theta \right]
Using p = 2qa:
V = \frac{1}{4 \pi \epsilon_0} \frac{p \cos \theta}{r^2}
And p \cdot \hat{r} = p \cos \theta
V(r) = \frac{1}{4 \pi \epsilon_0} \frac{p \cdot \hat{r}}{r^2} (Valid for r >> a).
For a point dipole p at the origin, this is exact.
Potential on the dipole axis (\theta = 0, \pi): V = \pm \frac{1}{4 \pi \epsilon_0} \frac{p}{r^2}
Positive sign for \theta = 0, negative sign for \theta = \pi.
Potential in the equatorial plane (\theta = \frac{\pi}{2}) is zero.

Contrasting Features of Dipole Potential vs. Single Charge Potential

Dipole potential depends on r and the angle between position vector r and dipole moment vector p. It is axially symmetric about p.
Electric dipole potential falls off as \frac{1}{r^2} at large distances, not as \frac{1}{r} (single charge).

Potential Due to a System of Charges

System of charges q1, q2, …, qn with position vectors r1, r2, …, rn.
Potential V1 at P due to charge q1: V1 = \frac{1}{4 \pi \epsilon0} \frac{q1}{r{1P}}
where r{1P} is the distance between q1 and P.
Similarly, V2 due to q2, V3 due to q3, etc.
By superposition principle, the potential V at P due to the total charge configuration is the algebraic sum:
V = V1 + V2 + … + Vn 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 with charge density \rho(r), divide it into small volume elements \Delta v carrying charge \rho \Delta v.
Determine potential due to each volume element and integrate over all such contributions.
For a uniformly charged spherical shell, the electric field outside the shell as if entire charge is concentrated at the centre, hence the potential outside the shell os;
V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r} (r >= R)
Electric field inside the shell is 0. Then, potential is constant inside the shell (as no work is done in moving a charge inside the shell), and equals its value at the surface V= q/4piE0*R

Example 2.2

Two charges 3 \times 10^{-8} C and -2 \times 10^{-8} C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
Let the origin O be at the location of the positive charge. The line joining the two charges is the x-axis; the negative charge is on the right side of the origin.
Let P be the required point on the x-axis where the potential is zero. If x is the x-coordinate of P, obviously x must be positive (as there is no possibility of potentials due to the two charges adding up to zero for x < 0).
If x lies between O and A, we have
\frac{1}{4 \pi \epsilon_0} \left[ \frac{3 \times 10^{-8}}{x} - \frac{2 \times 10^{-8}}{15 - x} \right] = 0
where x is in cm.
That is, \frac{3}{x} - \frac{2}{15 - x} = 0 which gives x = 9 cm.
If x lies on the extended line OA, the required condition is \frac{3}{x} - \frac{2}{x - 15} = 0 which gives x = 45 cm.
Electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. The formula for potential used in the calculation required choosing potential to be zero at infinity.

Example 2.3

Figures show the field lines of a positive and negative point charge respectively.
- Give the signs of the potential difference VP - VQ; VB - VA.
- Give the sign of the potential energy difference of a small negative charge between the points Q and P; A and B.
- Give the sign of the work done by the field in moving a small positive charge from Q to P.
- Give the sign of the work done by the external agency in moving a small negative charge from B to A.
- Does the kinetic energy of a small negative charge increase or decrease in going from B to A?
(a) As V \propto \frac{1}{r}, VP > VQ. Thus, (VP - VQ) is positive. Also VB is less negative than VA. Thus, VB > VA or (VB - VA) is positive.
(b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly, (P.E.)A > (P.E.)B and hence sign of potential energy differences is positive.
(c) In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative.
(d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive.
(e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from B to A.

Equipotential Surfaces

An equipotential surface is a surface with a constant value of potential at all points on the surface.
For a single charge q, the potential is given by V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r}.
V is constant if r is constant. Thus, equipotential surfaces of a single point charge are concentric spherical surfaces centered at the charge.
The electric field lines for a single charge q are radial lines starting from or ending at the charge.
The electric field at every point is normal to the equipotential surface passing through that point. This is true in general.
If the field were not normal, there would be a non-zero component along the surface.
Moving a unit test charge against the field component would require work, contradicting the definition of equipotential surface.
The electric field must be normal to the equipotential surface at every point.
Equipotential surfaces offer an alternative visual picture in addition to electric field lines around a charge configuration.
For a uniform electric field E along the x-axis, the equipotential surfaces are planes normal to the x-axis, i.e., planes parallel to the y-z plane.
Equipotential surfaces shown for a dipole and two identical positive charges.

Relation Between Field and Potential

Consider two closely spaced equipotential surfaces A and B with potential values V and V + \Delta V, where \Delta V is the change in V in the direction of the electric field E.
Let P be a point on the surface B. \Delta l is the perpendicular distance of the surface A from P.
Imagine that a unit positive charge is moved along this perpendicular from the surface B to surface A against the electric field. The work done in this process is |E|\Delta l.
This work equals the potential difference VA –VB.
|E|\Delta l = V – (V + \Delta V)= –\Delta V
|E|= -\frac{\Delta V}{\Delta l}
Since \Delta V is negative, \Delta V = - |\Delta V|.
E = −\frac{\Delta V}{\Delta l}
Two important conclusions:
- Electric field is in the direction in which the potential decreases steepest.
- Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.

Potential Energy of a System of Charges

Consider two charges q1 and q2 with position vectors r1 and r2.
Calculate the work done in building up this configuration (bringing charges from infinity to given locations).
Bring charge q1 from infinity to r1. No external field, so work done is zero.
This charge produces a potential V = \frac{1}{4 \pi \epsilon0} \frac{q1}{r_{1P}}
Work done in bringing charge q2 from infinity to r2 is q2 times the potential at r2 due to q1: *Work done on q2= \frac{1}{4 \pi \epsilon0} \frac{q1 q2}{r{12}} where r12 is the disntance between point 1 and 2.
Electrostatic force is conservative, so this work is stored as potential energy of the system.
*
The potential energy is: U = \frac{1}{4 \pi \epsilon0} \frac{q1 q2}{r{12}}
If q_2 was brought first, U would be the same due to path-independence.
Equation is true for any sign of q1 and q2.
If q1 q2 > 0, potential energy is positive (repulsive force).
If q1 q2 < 0, potential energy is negative (attractive force).
Generalized for three charges q1, q2, q3 located at r1, r2, r3.
Bring q1 from infinity to r1: no work.
Bring q2 from infinity to r2: work is \frac{1}{4 \pi \epsilon0} \frac{q1 q2}{r{12}}.
The potential is V = V1 + V2
V = \frac{1}{4 \pi \epsilon0} (\frac{q1}{r{1p}} + \frac{q2}{r_{2p}} )
Work done in bringing q3 from infinity to r3 is q3 \frac{1}{4 \pi \epsilon0} (\frac{q1}{r{13}} + \frac{q2}{r{23}} )
Total work done:
U = \frac{1}{4 \pi \epsilon0} \left( \frac{q1 q2}{r{12}} + \frac{q1 q3}{r{13}} + \frac{q2 q3}{r{23}} \right)
Because the force is conservative, The final expression for U is independent of assembly.
The potential energy is characteristic of configuration not the way it's achieved.

Example 2.4

Four charges are arranged at the corners of a square ABCD of side d.
- Find the work required to put together this arrangement.
- A charge q_0 is brought to the center E of the square, the four charges being held fixed at its corners. How much extra work is needed to do this?
(a) The work done depends on the final arrangement of the charges. Calculate work needed for one way of putting the charges at A, B, C and D.
Bring +q to A: zero work.
Bring -q to B: work is ( -q ( \frac{1}{4 \pi \epsilon0} \frac{q}{d} ) ) = \frac{-q^2}{4 \pi \epsilon0 d}.
Bring +q to C: work is (q ( \frac{1}{4 \pi \epsilon0} (\frac{q}{d \sqrt(2)} - \frac{q}{d} ) ) = \frac{q^2}{4 \pi \epsilon0 (d \sqrt(2)) } - \frac{q^2}{4 \pi \epsilon_0 d}
Bring -q to D: work is (-q \frac{1}{4 \pi \epsilon0} (\frac{q}{d} + \frac{q}{d \sqrt(2)} - \frac{q}{d} ) = \frac{q^2}{4 \pi \epsilon0 (d \sqrt(2) )}
Total work required is:
W = \frac{q^2}{4 \pi \epsilon0 d} ( -1 + \frac{1}{\sqrt(2)} - 1 - \frac{1}{\sqrt(2)} ) = \frac{-4q^2}{4 \pi \epsilon0 d} + \frac{2q^2}{\sqrt{2} 4 \pi \epsilon_0 d}
\frac{q^2}{4 \pi \epsilon_0 d} ( 4 + \sqrt(2) )
by definition, this is the total electrostatic energy of the charges. The students are encouraged to try with different assembling orders to show energy stays the same.
(b) Extra work necessary to bring a charge q0 to the point E is q0 \times (electrostatic potential at E due to the charges at A, B, C and D).
The electrostatic potential at E is zero since potential due to A and C is cancelled by that due to B and D. Hence, no work is required to bring any charge to point E.

Potential Energy in an External Field

Addresses potential energy of a charge q in a given field vs. potential energy of system of charges.
The external field E is not produced by the given charge(s).
The external sources may be known or unknown; what is specified is the electric field E or the electrostatic potential V due to the external sources.
Assume that the charge q does not significantly affect the sources producing the external field.
Consider determining the potential energy of a given charge q (and later, a system of charges) in the external field; not the potential energy of the sources producing the external electric field.
E and 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 the point P.
Work done in bringing a charge q from infinity to the point P in the external field is qV. This work is stored in the form of potential energy of q.
If the point P has position vector r, the potential energy of q at r in an external field is qV(r), where V(r) is the external potential at the point r.
Accelerating an electron ( q = -e = -1.6 \times 10^{-19} C ) by a potential difference of \Delta V = 1 volt yields energy of q \Delta V = 1.6 \times 10^{-19} J. Defined as 1 electron volt or 1eV.
1 eV=1.6 \times 10^{-19}J.

Potential Energy of a system of Two Charges in an External Field

What is the potential energy of a system of two charges q1 and q2 located at r1and r2, respectively, in an external field?
First, calculate the work done in bringing the charge q1 from infinity to r1. Work done in this step is q1 V(r1).
Next, we consider the work done in bringing q2 to r2. In this step, work is done not only against the external field E but also against the field due to q1.
Work done on q2 against the external field = q2 V (r2)
Work done on q2 against the field due to q1 = 1/(4 pi epsilon0 )(q1q2/r12) where r12 is the distance between q1 and q2 have made use of Equations (2.27) and (2.22).
By the superposition principle for fields, we add up the work done on q2 against the two fields (E and that due to q1):
Work done in bringing q2 to r2 = + q2 V (r2) + 1/(4 pi epsilon0 )(q1q2/r12) Thus
Potential energy of the system = the total work done in assembling the configuration.
Potential Energy = q1 V(r1) + q2 V (r2) + 1/(4 pi epsilon0 )(q1q2/r12)

Example 2.5

Determine the electrostatic potential energy of a system consisting of two charges 7 µC and –2 µC (and with no external field) placed at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively.
(b) How much work is required to separate the two charges infinitely away from each other?
*(c) Suppose that the same system of charges is now placed in an external electric field E = A (1/r 2); A = 9 × 105 NC–1 m2 . What would the electrostatic energy of the configuration be?
*Ue = (9 \times 10^9)(7 x 10-6)(-2 x 10-6)/0.18) = -0.7J.
*W = U2-U1 = 0 - U = 0 -(-0.7) =0.7J
(c) 7 \times 10^{-6} A/0.09 + (-2 \times 10^{-6} A/0.09) = 70 -20 -0.7 =49.3 J.

Potential Energy of a Dipole in an External Field

Consider a dipole with charges q1 = +q and q2 = -q placed in a uniform electric field E.
In a uniform electric field, the dipole experiences no net force; but experiences a torque \tau = p \times E which will tend to rotate it (unless p is parallel or anti-parallel to E).
Suppose an external torque \tau{\text{ext}} is applied, neutralizing this torque, rotating dipole in plane of paper, from \theta0 to \theta1 at an infinitesimal angular speed and no angular accelerating. The amount of work done by the external torque W = \int{\theta0}^{\theta1} \tau \ext d\theta = \int{\theta0}^{\theta_1} p E \sin \theta d\theta = p E (\cos \theta0 - \costheta1 )
This work is stored as the potential energy of the system. We can then associate potential energy with an inclination of the dipole. Similar to other potential energies, there is a freedom in choosing the angle where the potential energy U is taken to be zero. A natural choice is to take
U = - p \cdot E = −pE \cos\theta. Potential can be alternatvely unsterstood from:

( ) 2 1 2 [ ] 4 2 q U q V V = r r where r and 1 and r2 are respective position vectors of
( r1 ) - ( r2) = E *a *costheta.
*In this case, the work done against the external field E in bringing +q and – q are equal and opposite and cancel out, i.e., q [V (r1) – V (r2)]=0.

Example 2.6

A molecule of a substance has a permanent electric dipole moment of magnitude 10–29 C m. A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude 106 V m–1. The direction of the field is suddenly changed by an angle of 60º. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarisation of the sample.

Sol:
Dipole moment of each molecules = 10–29 C m
1 mole of the substance contains 6 × 1023 molecules
total dipole moment of all the molecules = 6 × 1023 × 10–29 C m = 6 × 10–6 C m

Initial potential energy = –pE cos
= –6×10–6×106 cos 0° = –6 J
Final potential energy (when = 60°)= –6 × 10–6 × 106 cos 60° = –3 J

Change in potential energy = -3J -(-6J) = 3J
energy released by the substance in the form of heat is 3 J

Electrostatics of Conductors

Conductors contain mobile charge carriers (electrons in metallic conductors, positive and negative ions in electrolytic conductors).
In a metal, outer electrons are free to move within the metal but not free to leave.
Free electrons form a gas, colliding with each other and with ions, moving randomly.
In an external electric field, they drift against the direction of the field.
Positive ions remain in fixed positions.

Important Results Regarding Electrostatics of Conductors

Inside a conductor, electrostatic field is zero
Result means that a conductor has free electrons. As long as electric field is not zero, the free charge carriers would experience force and drift. In the static situation, the free charges have so distributed themselves that the electric field is zero everywhere inside.
At the surface of a charged conductor, electrostatic field must be normal to the