Electrostatic Potentials

Question 1

What is the equation for the electric field in terms of the electrostatic potential?

Answer 1

E = -∇ψ, where ψ is the electrostatic potential

Question 2

What is meant by the negative sign in front of the electrostatic potential in words?

Answer 2

The work down when a positive charge moves from a high to a low potential.

Question 3

What is Poisson’s equation?

Answer 3

∇^2 ψ = -ρ/ε0

Question 4

What is Laplace’s equation?

Answer 4

∇^2 ψ = 0

Question 5

If we consider a potential in the x-y plane which takes the form ψ = V0*cos(kx), how do we find how the potential varies with z?

Answer 5

• Laplace’s equation becomes d^2ψ/dx^2 + d^2ψ/dz^2 = 0, as ψ is not dependent on y
• Trial solution ψ = cos(kx)exp(az), sub this in, and rearrange to get equation for ψ in terms of x and z (by finding equation for a)

Question 6

If we have a potential ψ = -E0rcosθ for a dielectric cylinder oriented along z-axis with field running in x-direction, what does Laplace’s equation equal in cylindrical polars?

Answer 6

1/rd/dr(rdψ/dr) + 1/r^2 *d^2ψ/dθ^2 = 0

Question 7

What trial solution do we use to solve this version of Laplace’s equation, and what do we find?

Answer 7

Use trial solution of ψ = f(r)cosθ, sub this in, and find that rdr/dr(r*df/dr) - f = 0, so general solution is f(r) = Ar+Br^-1

Question 8

What do we need to do to find the potential inside and outside of the dielectric cylinder?

Answer 8

Try solution for ψ(r<=a) and ψ(r>=a), using A1, B1 and A2, B2.

Question 9

What are the two equations we need to solve for ψ? What can we find straight away?

Answer 9

ψ(r<=a) = (A1r+B1/r)cosθ, ψ(r>=a) = (A2r+B2/r)cosθ. We straight away know that A2=-E0, since the B2 terms dies away with r, and to avoid infinities as r->0, we need B1 = 0, so left with B2 and A1

Question 10

What is the equation for D?

Answer 10

D = ε0εrE, with εr = 1 outside cylinder and E = -∇ψ

Question 11

Which boundary conditions do we look at to solve the potential inside and outside the cylinder?

Answer 11

r=a and θ = 0

Question 12

What condition do we get from the boundary conditions and the fact that ∇.D and ∇.B = 0?

Answer 12

-εrdψ(r<=a)/dr at r=a = -dψ(r>=a)/dr at r=a, leading to -εrA1 = E0+B2/a^2, and since εr = 1, we can multiply through by a to get final equation

Question 13

What is the equation for A1 equal to?

Answer 13

A1 = -2/(εr+1) *E0

Question 14

What is the equation for B2 equal to?

Answer 14

B2 = (εr-1)/(εr+1) a^2E0

Question 15

What do we deduce from the equations for A1 and B2?

Answer 15

Field strength inside the cylinder is a factor of 2/(εr+1) times the field outside.