Electrostatic Boundary Value problems

1) Coulomb’s law 2) Gauss’s law

Application of these techniques for determining electric field is possible when the charge distribution is known. In most practical situations, this is certainly not the case.

In practical electrostatic problems, electrostatic conditions (charge or potential) at some boundaries are known, and it is desired to find E and V throughout the region. These are referred to as Boundary Value Problems.

∇.𝐷𝐷 = ∇. 𝜀𝜀𝜀𝜀 = 𝜌𝜌

Laplace’s/Poisson’s equation:

𝜀𝜀 = −∇.𝑉𝑉

∇. (−𝜀𝜀∇𝑉𝑉) = 𝜌𝜌 ∇2 𝑉𝑉 = −𝜌𝜌𝜀𝜀

∇2 𝑉𝑉 = 0 (homogeneous region, ε is constant)

Laplace’s equation is of primary importance in solving electrostatic problems.

If a solution of Laplace’s equation satisfies a given set of boundary conditions, is this the only possible solution? Let us assume that there are two solutions V1 and V2. Both of them satisfy Laplace’s equation with the same set of prescribed boundary conditions.

Uniqueness Theorem

∇2 𝑉𝑉1 = 0

∇2 𝑉𝑉2 = 0 V1 = V2 on the boundary

Say, Vd = V2 – V1

∇2 𝑉𝑉𝑑𝑑 = ∇2 𝑉𝑉2 − ∇2 𝑉𝑉1 = 0

Also, Vd =0 on the boundary.

Let 𝐴𝐴 = 𝑉𝑉𝑑𝑑 .∇𝑉𝑉𝑑𝑑

∇.𝐴𝐴 = ∇. (𝑉𝑉𝑑𝑑∇𝑉𝑉𝑑𝑑) = 𝑉𝑉𝑑𝑑∇2 𝑉𝑉𝑑𝑑 + (∇𝑉𝑉𝑑𝑑)2 (identity)

∇.𝐴𝐴 = (∇𝑉𝑉𝑑𝑑)2

From Divergence theorem, ∫ ∇.𝐴𝐴 𝑑𝑑𝑑𝑑 = ∮𝐴𝐴.𝑑𝑑𝑑𝑑𝑑𝑑𝑣𝑣𝑣𝑣

∫ (∇𝑉𝑉𝑑𝑑)2𝑑𝑑𝑑𝑑 = ∮ 𝑉𝑉𝑑𝑑∇Vd.ds���𝑑𝑑𝑑𝑑𝑣𝑣𝑣𝑣

Since Vd = 0, on the boundary, surface integral equates to zero. So the volume integral must also vanish.

∴ �(∇𝑉𝑉𝑑𝑑)2

𝑑𝑑𝑣𝑣𝑣𝑣

𝑑𝑑𝑑𝑑 = 0

Since the integration is always positive, ∇𝑉𝑉𝑑𝑑 = 0

therefore, Vd = V2 – V1 = constant everywhere in the volume. But Vd =0 on the boundary, therefore Vd = 0 everywhere. Therefore V1 & V2 cannot be different solutions to the same problem.

Uniqueness Theorem

Boundary Value Problem is uniquely defined by: -

: - If a solution to Laplace equation can be found that satisfies the boundary conditions, then the solution is unique. The theorem can be similarly shown to hold for Poisson’s equation as well.

1) The appropriate differential equation 2) Solution region 3) Prescribed boundary conditions

Partial Differential Equations

The Laplace is a PDE. A host of physics problems involve solution of PDE’s. PDE’s which occur in physics are mostly second order.

General PDE in 2 dimensions:

𝑎𝑎𝜕𝜕2𝑉𝑉𝜕𝜕𝜕𝜕2 + 𝑏𝑏

𝜕𝜕2𝑉𝑉𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕

+ 𝑐𝑐𝜕𝜕2𝑉𝑉𝜕𝜕𝜕𝜕2 + 𝑑𝑑

𝜕𝜕𝑉𝑉𝜕𝜕𝜕𝜕

+ 𝑒𝑒𝜕𝜕𝑉𝑉𝜕𝜕𝜕𝜕

+ 𝑓𝑓𝑉𝑉 + 𝑔𝑔 = 0

b2 < ac: elliptic e.g. Laplace’s equation, 𝜕𝜕2𝑉𝑉𝜕𝜕𝜕𝜕 2 + 𝜕𝜕2𝑉𝑉

𝜕𝜕𝜕𝜕 2 = 0

b2 > ac: hyperbolic e.g. Wave equation, 𝜕𝜕2𝑉𝑉𝜕𝜕𝜕𝜕 2 −

1𝑐𝑐2

𝜕𝜕2𝑉𝑉𝜕𝜕𝜕𝜕 2 = 0

b2 = ac: parabolic equation e.g. diffusion equation, D 𝜕𝜕2𝑉𝑉𝜕𝜕𝜕𝜕 2 −

𝜕𝜕𝑉𝑉𝜕𝜕𝜕𝜕

= 0

Generally elliptic equations have boundary conditions which are specified around a closed boundary. Hyperbolic and Parabolic equations have at least one open boundary. The boundary conditions for at least one variable are specified at one end of the system. The wave equation and the diffusion equation contain a time variable. There is a set of initial conditions at a particular time. These properties are related to the fact that an ellipse is a closed object, whereas hyperbola and parabola are open.

Important point:

General procedures for solution of Laplace’s equation:

1) Direct Integration (when V is a function of one variable) 2) Separation of variables (V is a function of more than one variable)

The solution at this point is expressed in terms of unknown integration constants.

Apply the boundary conditions to determine a unique solution for V.

Having obtained V, 𝜀𝜀 = −∇.𝑉𝑉 and D = εE.

If necessary, the charge Q on a conductor can be computed, 𝑄𝑄 = ∫𝜌𝜌𝑑𝑑𝑑𝑑𝑑𝑑, where 𝜌𝜌𝑑𝑑 = 𝐷𝐷𝑛𝑛 , 𝐷𝐷𝑛𝑛 being the normal component of D at the conductor.

Capacitance may be computed: C=Q/V

A photocopying machine is an important application of electrostatics. The surface of the photoconductor is initially charged uniformly. When light from the document to be copied is focussed on the photoconductor, the charges on the lower surface combine with those on the upper surface to neutralise each other. The image is developed by pouring a charged black powder over the surface of the photoconductor. The electric field attracts the charged powder, which is later transferred to paper and melted to form a permanent image. We want to determine the electric field below and above the surface of the photoconductor.

Example 1: sadiku pg. 205

𝜌𝜌𝑑𝑑 = 0 , therefore we need to solve laplace’s equation. Also the potential depends only on 𝜕𝜕.

∇2𝑉𝑉 = 0

𝜕𝜕2𝑉𝑉𝜕𝜕𝜕𝜕 2 = 0

𝜕𝜕𝑉𝑉𝜕𝜕𝜕𝜕

= 𝐵𝐵

V = Ax + B

Let, V1 = A1x + B1, x > a

V2 = A2x + B2, x< a

V1(x=d) = 0, V2(x=0) = 0

At x = a, V1 = V2 and D1n – D2n = ρs --- boundary conditions

0 = A1d + B1, => B1 = - A1d

0 = 0 + B2, => B2 = 0

A1a + B1 = A2a at x=a

D = εE = −𝜀𝜀∇𝑉𝑉

𝜌𝜌𝑑𝑑 = 𝐷𝐷1𝑛𝑛 − 𝐷𝐷2𝑛𝑛 = 𝜀𝜀1𝜀𝜀1𝑛𝑛 − 𝜀𝜀2𝜀𝜀2𝑛𝑛 = −𝜀𝜀1𝑑𝑑𝑉𝑉1𝑑𝑑𝜕𝜕

+ 𝜀𝜀2𝑑𝑑𝑉𝑉2𝑑𝑑𝜕𝜕

= −𝜀𝜀1𝐴𝐴1 + 𝜀𝜀2𝐴𝐴2

Solving for A1 and A2, we get

𝜀𝜀1 = −𝐴𝐴1𝑎𝑎�𝜕𝜕

𝜀𝜀2 = −𝐴𝐴2𝑎𝑎�𝜕𝜕

Determine the potential function for the region inside the rectangular trough of infinite length whose cross section is shown.

Example 2:

∇2𝑉𝑉 = 0

Boundary conditions:

V(x=0, 0 ≤ y ≤ a) = 0 V(x=b, 0 ≤ y ≤ a) = 0 V (0 ≤ x ≤ b, y = 0) = 0 V (0 ≤ x ≤ b, y = a) = Vo

Method of separation of variables:

Let V(x,y) = X(x).Y(y)

X”Y + Y”X = 0

YY

XX ""

=−

Since LHS is a function of x only and RHS is a function of y only, this equality holds only if the ratios are constant.

Let λ==−YY

XX ""

, λ is called the separation constant

The variables have been separated as follows –

X” + λx = 0 & Y” + λy = 0

Boundary conditions:

V (0,y) = X(0)Y(y) = 0 X(0) = 0

V(b,y) = X(b)Y(y) = 0 X(b) = 0

V (x,y) = X(x)Y(0) = 0 Y(0) = 0

V(x,a) = X(x)Y(a) = Vo

Let us consider all possible values of λ

Let λ = 0

Case A:

X” = 0 =>X= Ax + B

X(x=0) = 0 => B = 0

X(x=b) = 0 => A = 0

X(x) = 0 i.e. V = 0

This becomes a trivial solution. Since λ ≠ 0

Let λ < 0, say λ = - α2

Case B:

X” - α2 X = 0

(D2 - α2 )X = 0, D = d/dx

DX = ± αX

If, DX = αX

XdxdX α= or, ∫ ∫= dx

XdX α or, ln(X) = αx + ln(A1)

or xeAX α1=

If DX = - αX xeAX α2=

Total solution, xx eAeAxX αα −+= 21)(

𝑋𝑋(𝜕𝜕) = 𝐴𝐴1(cosh𝛼𝛼𝜕𝜕 + sinh𝛼𝛼𝜕𝜕) + 𝐴𝐴2(cosh𝛼𝛼𝜕𝜕 − sinh𝛼𝛼𝜕𝜕)

𝑋𝑋(𝜕𝜕) = 𝐵𝐵1 cosh𝛼𝛼𝜕𝜕 + 𝐵𝐵2 sinh𝛼𝛼𝜕𝜕

where, B1 = A1 + A2 and B2 = A1 – A2

X(x=0) = 0 B1=0

X(x=b)=0 B2 =0

Therefore X(x) =0. Also a trivial solution for 𝜆𝜆 ≮ 0

λ>0, say λ = β2

Case C:

X” + β2X = 0

(D2 + β2)X = 0

DX = ± jβX

As before, 𝑋𝑋(𝜕𝜕) = 𝐶𝐶0𝑒𝑒𝑗𝑗𝑗𝑗𝜕𝜕 + 𝐶𝐶1𝑒𝑒−𝑗𝑗𝑗𝑗𝜕𝜕

where, 𝑒𝑒𝑗𝑗𝑗𝑗𝜕𝜕 = cos𝑗𝑗𝜕𝜕 + 𝑗𝑗 sin𝑗𝑗𝜕𝜕 and 𝑒𝑒−𝑗𝑗𝑗𝑗𝜕𝜕 = cos𝑗𝑗𝜕𝜕 − 𝑗𝑗 sin𝑗𝑗𝜕𝜕

𝑋𝑋(𝜕𝜕) = 𝑔𝑔0 cos𝑗𝑗𝜕𝜕 + 𝑔𝑔1 sin𝑗𝑗𝜕𝜕

𝑔𝑔0 = 𝐶𝐶0 + 𝐶𝐶1 and 𝑔𝑔1 = 𝐶𝐶0 − 𝑗𝑗𝐶𝐶1

X(x=0) = 0 g0 = 0

X(x=b) = 0 g1sin βb = 0, If g1 ≠ 0, sin βb = 0 = sin nπ , therefore β = nπ/b, where n= 1,2,3,........

sinh x = 0, only when x=0

sin x = 0 at infinite no. of pts.

If n = 0, β = 0, gives a trivial solution.

If n = -1, -2, ..... is not needed to be considered since λ = β2 has the same value whether β is (+) or (-)

𝑋𝑋𝑛𝑛(𝜕𝜕) = 𝑔𝑔𝑛𝑛 sin 𝑛𝑛𝑛𝑛𝜕𝜕𝑏𝑏

for a given value of n

𝜆𝜆 = 𝑗𝑗2 = 𝑛𝑛2𝑛𝑛2

𝑏𝑏2

Let us now find Y(y)

𝑌𝑌" − 𝑗𝑗2𝑌𝑌 = 0

𝑌𝑌(𝜕𝜕) = ℎ0 cosh𝑗𝑗𝜕𝜕 + ℎ1 sinh𝑗𝑗𝜕𝜕

Y(y=0) = 0 ℎ0 = 0

𝑌𝑌(𝜕𝜕) = ℎ𝑛𝑛 sinh 𝑛𝑛𝑛𝑛𝜕𝜕𝑏𝑏

∴ 𝑉𝑉𝑛𝑛(𝜕𝜕,𝜕𝜕) = 𝑔𝑔𝑛𝑛ℎ𝑛𝑛 sin 𝑛𝑛𝑛𝑛𝜕𝜕𝑏𝑏

sinh 𝑛𝑛𝑛𝑛𝜕𝜕𝑏𝑏

Thus there are many possible solutions of V for various values. If V1, V2, V3,...... are solutions of Laplace’s equations, then by superposition theorem,

then V = e1V1 + e2V2 + .......+ enVn, is also a solution.

𝑉𝑉(𝜕𝜕,𝜕𝜕) = ∑ 𝐶𝐶𝑛𝑛 sin 𝑛𝑛𝑛𝑛𝜕𝜕𝑏𝑏

sinh 𝑛𝑛𝑛𝑛𝜕𝜕𝑏𝑏

∞𝑛𝑛=1 where 𝐶𝐶𝑛𝑛 = 𝑔𝑔𝑛𝑛ℎ𝑛𝑛

Cn is to be determined from boundary conditions

𝑉𝑉(𝜕𝜕,𝜕𝜕 = 𝑎𝑎) = 𝑉𝑉0 = ∑ 𝐶𝐶𝑛𝑛 sin 𝑛𝑛𝑛𝑛𝜕𝜕𝑏𝑏

sinh 𝑛𝑛𝑛𝑛𝜕𝜕𝑏𝑏

∞𝑛𝑛=1

This is a Fourier Series expansion of Vo

∫ 𝑉𝑉0 sin𝑚𝑚𝑛𝑛𝜕𝜕𝑏𝑏𝑑𝑑𝜕𝜕 =𝑏𝑏

0 ∑ 𝐶𝐶𝑛𝑛 sin 𝑛𝑛𝑛𝑛𝑎𝑎𝑏𝑏

∞𝑛𝑛=1 ∫ sin𝑚𝑚𝑛𝑛𝜕𝜕

𝑏𝑏𝑏𝑏

0 sin 𝑛𝑛𝑛𝑛𝜕𝜕𝑏𝑏𝑑𝑑𝜕𝜕

� sin𝑚𝑚𝜕𝜕 sin𝑛𝑛𝜕𝜕 𝑑𝑑𝜕𝜕 = 0 𝑚𝑚 ≠ 𝑛𝑛𝑛𝑛

0

� sin𝑚𝑚𝜕𝜕 sin𝑛𝑛𝜕𝜕 𝑑𝑑𝜕𝜕 =𝑛𝑛2

𝑚𝑚 = 𝑛𝑛𝑛𝑛

0

Therefore, all terms ∫ sin𝑚𝑚𝑛𝑛𝜕𝜕𝑏𝑏

𝑏𝑏0 sin 𝑛𝑛𝑛𝑛𝜕𝜕

𝑏𝑏𝑑𝑑𝜕𝜕 = ∫ 𝑑𝑑𝑠𝑠𝑛𝑛2 𝑛𝑛𝑛𝑛𝜕𝜕

𝑏𝑏𝑏𝑏

0 𝑑𝑑𝜕𝜕 vanish on the RHS unless m = n.

� 𝑉𝑉0 sin𝑛𝑛𝑛𝑛𝜕𝜕𝑏𝑏

𝑑𝑑𝜕𝜕 =𝑏𝑏

0𝐶𝐶𝑛𝑛 sinh

𝑛𝑛𝑛𝑛𝑎𝑎𝑏𝑏

� 𝑑𝑑𝑠𝑠𝑛𝑛2 𝑛𝑛𝑛𝑛𝜕𝜕𝑏𝑏

𝑑𝑑𝜕𝜕𝑏𝑏

0

−𝑉𝑉0.𝑏𝑏𝑛𝑛𝑛𝑛

. �cos𝑛𝑛𝑛𝑛𝜕𝜕𝑏𝑏 �

0

𝑏𝑏= 𝐶𝐶𝑛𝑛 sinh

𝑛𝑛𝑛𝑛𝑎𝑎𝑏𝑏

.12

.� (1 − cos2𝑛𝑛𝑛𝑛𝜕𝜕𝑏𝑏

)𝑑𝑑𝜕𝜕𝑏𝑏

0

𝑉𝑉0𝑏𝑏𝑛𝑛𝑛𝑛

(1 − cos𝑛𝑛𝑛𝑛) = 𝐶𝐶𝑛𝑛 sinh𝑛𝑛𝑛𝑛𝑎𝑎𝑏𝑏

.𝑏𝑏2

𝐶𝐶𝑛𝑛 sinh𝑛𝑛𝑛𝑛𝑎𝑎𝑏𝑏

=2𝑉𝑉0

𝑛𝑛𝑛𝑛(1 − cos𝑛𝑛𝑛𝑛)

𝐶𝐶𝑛𝑛 sinh𝑛𝑛𝑛𝑛𝑎𝑎𝑏𝑏

= 4𝑉𝑉0

𝑛𝑛𝑛𝑛, 𝑛𝑛 = 1,3,5, … … … … … …

𝐶𝐶𝑛𝑛 sinh𝑛𝑛𝑛𝑛𝑎𝑎𝑏𝑏

= 0, 𝑛𝑛 = 2,4,6, … … … … … …

𝐶𝐶𝑛𝑛 =4𝑉𝑉0

𝑛𝑛𝑛𝑛 sinh𝑛𝑛𝑛𝑛𝑏𝑏 𝑠𝑠𝑓𝑓 𝑛𝑛 𝑠𝑠𝑑𝑑 𝑣𝑣𝑑𝑑𝑑𝑑

𝐶𝐶𝑛𝑛 = 0 𝑠𝑠𝑓𝑓 𝑛𝑛 𝑠𝑠𝑑𝑑 𝑒𝑒𝑑𝑑𝑒𝑒𝑛𝑛

Complete solution

𝑉𝑉(𝜕𝜕,𝜕𝜕) =4𝑉𝑉0

𝑛𝑛�

sin𝑛𝑛𝑛𝑛𝜕𝜕𝑏𝑏 sinh𝑛𝑛𝑛𝑛𝜕𝜕𝑏𝑏𝑛𝑛 sinh𝑛𝑛𝑛𝑛𝑎𝑎𝑏𝑏

∞

𝑛𝑛=1,3,5…

Check the solution at the boundaries.

To determine the potential for each point (x,y) in the trough, we take the first few term of the convergent infinite series.

Computation of Capacitance:

𝐶𝐶 =𝑄𝑄𝑉𝑉

=𝜀𝜀 ∮𝜀𝜀�⃑ .𝑑𝑑𝑑𝑑����⃑

∫ 𝜀𝜀�⃑ .𝑑𝑑𝑣𝑣���⃑

For any two conductor capacitor, C may be determined.

1) Gauss’s Law: Assume Q, and find V in terms of Q 2) Laplace’s equation: Assume V, determine Q in terms of V

Determine E using Gauss’s or Coulomb’s law:-

𝑉𝑉 = −∫𝜀𝜀�⃑ .𝑑𝑑𝑣𝑣���⃑ 𝐶𝐶 = 𝑄𝑄𝑉𝑉

Parallel Plate Capacitor:

Each Plate has area S. Charges are uniformly distributed.

𝜌𝜌𝑑𝑑 =𝑄𝑄𝑆𝑆

Ideal capacitor: distance between plates << dimensions of plate. So, fringing can be ignored.

1st method:

𝐷𝐷 = −𝜌𝜌𝑑𝑑𝜕𝜕�

𝜀𝜀 =𝜌𝜌𝑑𝑑𝜀𝜀

(−𝜕𝜕�) = −𝑄𝑄𝜀𝜀𝑆𝑆𝜕𝜕�

𝑉𝑉 = −�𝜀𝜀�⃑ .𝑑𝑑𝑣𝑣���⃑1

2

= −� �−𝑄𝑄𝜀𝜀𝑆𝑆𝜕𝜕�.𝑑𝑑𝜕𝜕𝜕𝜕�� =

𝑄𝑄𝑑𝑑𝜀𝜀𝑆𝑆

𝑑𝑑

0

𝐶𝐶 =𝑄𝑄𝑉𝑉

=𝜀𝜀𝑆𝑆𝑑𝑑

𝜀𝜀𝑛𝑛𝑒𝑒𝐸𝐸𝑔𝑔𝜕𝜕,𝑊𝑊𝜀𝜀 =12� 𝜀𝜀𝜀𝜀2𝑑𝑑𝑑𝑑 =

12� 𝜀𝜀

𝑄𝑄2

𝜀𝜀2𝑆𝑆2 𝑑𝑑𝑑𝑑 =𝜀𝜀𝑄𝑄2𝑆𝑆𝑑𝑑2𝜀𝜀2𝑆𝑆2 =

𝑄𝑄2𝑑𝑑2𝜀𝜀𝑆𝑆

𝑑𝑑𝑣𝑣𝑣𝑣

=𝑄𝑄2

2𝐶𝐶=

12𝑄𝑄𝑉𝑉

𝑑𝑑𝑣𝑣𝑣𝑣

Problem:

Conducting Spherical shells are maintained at a voltage difference of Vo.

V(r=b) = 0

V(r=a) = Vo

Determine V and E in the region between shells. V depends only on r.

Laplace’s equation: ∇2= 0

1𝐸𝐸2

𝑑𝑑𝑑𝑑𝐸𝐸�𝐸𝐸2 𝑑𝑑𝑑𝑑

𝑑𝑑𝐸𝐸� = 0

𝑑𝑑𝑑𝑑𝐸𝐸�𝐸𝐸2 𝑑𝑑𝑑𝑑

𝑑𝑑𝐸𝐸� = 0

�𝐸𝐸2 𝑑𝑑𝑑𝑑𝑑𝑑𝐸𝐸� = 𝐴𝐴

2nd method (Laplace’s equation):

𝑑𝑑𝑑𝑑𝑑𝑑𝐸𝐸

=𝐴𝐴𝐸𝐸2

𝑉𝑉 = −𝐴𝐴𝐸𝐸

+ 𝐵𝐵

r = b, V = 0 𝐵𝐵 = 𝐴𝐴𝑏𝑏

r = a, V = Vo 𝑉𝑉0 = 𝐴𝐴(1𝑏𝑏− 1

𝑎𝑎)

∴ 𝐴𝐴 = 𝑉𝑉0

1𝑏𝑏 −

1𝑎𝑎

𝑉𝑉 = 𝑉𝑉0

1𝐸𝐸 −

1𝑏𝑏

1𝑎𝑎 −

1𝑏𝑏

𝜀𝜀 = −∇.𝑉𝑉 = −𝑑𝑑𝑑𝑑𝑑𝑑𝐸𝐸

�̂�𝐸 = −𝐴𝐴𝐸𝐸2 �̂�𝐸 =

𝑉𝑉0

𝐸𝐸2(1𝑎𝑎 −

1𝑏𝑏)�̂�𝐸

𝑄𝑄 = �𝜀𝜀𝜀𝜀.𝑑𝑑𝑑𝑑 = � �𝜀𝜀𝑣𝑣𝜀𝜀𝐸𝐸𝑉𝑉𝑣𝑣

𝐸𝐸(1𝑎𝑎 −

1𝑏𝑏)

2𝑛𝑛

𝜑𝜑=0

𝐸𝐸2 sin𝜃𝜃𝑑𝑑𝜑𝜑𝑑𝑑𝜃𝜃 =4𝑛𝑛𝜀𝜀𝑣𝑣𝜀𝜀𝐸𝐸𝑉𝑉𝑣𝑣

1𝑎𝑎 −

1𝑏𝑏

𝑛𝑛

𝜃𝜃=0

𝐶𝐶 =𝑄𝑄𝑉𝑉

=4𝑛𝑛𝜀𝜀

1𝑎𝑎 −

1𝑏𝑏

1st method (gauss’s law):

𝑄𝑄 = 𝜀𝜀 �𝜀𝜀.𝑑𝑑𝑑𝑑 = 𝜀𝜀𝜀𝜀𝐸𝐸 . 4𝑛𝑛𝐸𝐸2

𝜀𝜀 =𝑄𝑄

4𝑛𝑛𝜀𝜀𝐸𝐸2 �̂�𝐸

𝑉𝑉 = −�𝜀𝜀�⃑ .𝑑𝑑𝑣𝑣���⃑1

2

= −��𝑄𝑄

4𝑛𝑛𝜀𝜀𝐸𝐸2 �̂�𝐸�𝑎𝑎

𝑏𝑏

.𝑑𝑑𝐸𝐸. 𝐸𝐸� =𝑄𝑄

4𝑛𝑛𝜀𝜀�

1𝑎𝑎−

1𝑏𝑏�

𝐶𝐶 =𝑄𝑄𝑉𝑉

=4𝑛𝑛𝜀𝜀

1𝑎𝑎 −

1𝑏𝑏