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![A point charge Q [C] is located at a point in space. The charge is surrounded by two spherical layers of materials as shown b](http://img.homeworklib.com/images/4507470d-0136-4793-9b84-9ac00dd51caa.png?x-oss-process=image/resize,w_560)
Homework Answers
given point charge Q
permittivity of the materials, epsilon1, epsilon2
epsilon2 > epsilon1
a. considering a spherical gaussean surface around the charge at
radius
let r < R1
then
from gauss law
electric flux density = phi(r)
phi(r) = Q/4*pi*r^2
for R1 <= r <= R2
then again
phi(r) = Q/4*pi*r^2
for R2 <= r <= R3
then again
phi(r) = Q/4*pi*r^2
for R3 <= r
phi(r) = Q/4*pi*r^2
b. now, electric field is given by E(r)
for r <= R1
E(r) = Q/4*pi*r^2*epsilon0
now
E = -dV/dr
hence
V(r) = Q/4*pi*r*epsilon0
for R1<= r <= R2
V(r) = Q/4*pi*R1*epsilon1 - Q/4*pi*r*epsilon1 +
Q/4*pi*R1*epsilon0
for R2 <= r <= R3
V(r) = Q/4*pi*R1*epsilon1 - Q/4*pi*R2*epsilon +
Q/4*pi*R1*epsilon0 + Q/4*pi*R2*epsilon2 - Q/4*pi*r*epsilon2
for r >= R3
V(r) = Q/4*pi*R1*epsilon0 + Q/4*pi*R1*epsilon1 -
Q/4*pi*R2*epsilon1 + Q/4*pi*R2*epsilon2 - Q/4*pi*R3*epsilon2 +
Q/4*pi*R3*epsilon - Q/4*pi*epsilon*r
c. electric flux density graph is as under
electric field intensity graph is as under
d. in region R2
electric field is given by
E(r) = Q/4*pi*r^2*epsilon1
now
u = 0.5*epsilon1*E^2 [ energy density]
hence energy
U = 0.5*epsilon1*integral[E(r)^2 4*pi*r^2*dr]
U = integral[Q^2*dr/8*pi*r^2*epsilon1]
U = Q^2/8*pi*R1*epsilon1 - Q^2/8*pi*r*epsilon1
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