Question

A spherical capacitor has a spherical inner plate with radius a and outer plate with radius b. The charge on the inner plate is +Q and on the outer plate it is -Q. We have filled a cone shaped region of angle θ (0 ≤ θ ≤ π) with a dielectric with constant κ. The dielectric fills the entire volume between the two spheres inside the cone. You may neglect any fringing effects between the dielectric and the vacuum (dielectric constant = 1) that is in the rest of the capacitor. (Hint: If you put a dielectric part way into the space between two plates it can be considered as two capacitors in parallel.) (Note: The green shaded region is just to help illustrate where the cone is.)

a) Let θ = 0 so that we’re just working with the standard spherical capacitor. Using Gauss’ Law, find the electric field E, potential difference ∆V, and capacitance C for this case.

b) Show that the area of the darkly shaded region of the outer sphere is A = 2πb²(1-cosθ).

c) Ignoring any fringing, calculate the capacitance of the cone shaped region. (Hint: you’ll want to think of this as a fraction of a spherical capacitor and use the ratio of the areas.)

d) Repeat what you did in (c) for the region outside of the cone (i.e. the rest of the sphere), using the same trick.

e) Since we’re considering this as two capacitors in parallel, find the total capacitance of this system in terms of θ, a, b, κ and fundamental constants.

f) Find the total energy stored in the capacitor, U, when it has charge Q on it as a function of θ.

g) For what value of θ is the energy stored at a maximum? For what value is it at a minimum? What does this tell us about how dielectrics can be useful in electronic devices if we want to minimize energy consumption?

Answer 1

Given spherical capacitor

inner plate radius = a

outer plate radius = b

CHarge = Q

dielectric constant = k'

half angle of the cone shape = theta

a. for theta = 0, we have normal spherical capacitor

using gauss law for a spherical gaussean surface at radius r, r < b, r > a

electric field inside this region = E

E*4*pi*r^2 = Q/epsilon

E = kQ/r^2 [ where k is coloumbs constant]

now, E = -dV/dr

so, dV = -kQ*dr/r^2

so integrating from r = b to r = a

V = kQ(1/a - 1/b) = kQ(b - a)/ab

now for a capacitor with capacitance C

Q = CV

hence C = Q/V = Qab/kQ(b - a) = ab/k(b - a) [ where k is coloumbs constant]

b. if we consider a ring on the outer surface at half angle theta

thickness of ring, dx = b*d(theta)

area of this ring = 2*pi*b^2*sin(theta)*d(theta)

total area of the shaded region, integrate dA form theta = 0 to theta = theta

so, A = 2*pi*b^2*(cos(0) - cos(theta)) = 2*pi*b^2*(1 - cos(theta))

c. capacitance of this shaded area be C'

then C'/C = A/A'

where A' is area of the outer sphere = 4*pi*b^2

so, C' = 2*pi*b^2*(1 - cos(theta)) *C/4*pi b^2 = (1 - cos(theta))*ab/2k(b - a)

d. for region outside the cone, let capacitance be C"

C"/C = (4pi*b^2 - 2*pi*b^2*(1 - cos(theta)))/4*pi*b^2 = (1 + cos(theta)))/2

C" = (1 + cos(theta))*ab/2kQ(b - a)

e. as these two capacitros are in parallel, and the smaller one has a dielectric of dielectric constant k'

net capacitance = k'*(1 - cos(theta))*ab/2k(b - a) + (1 + cos(theta))*ab/2kQ(b - a)