Two thin, isolated, concentric conducting spheres of radii 1.10 cm and 7.70 cm,

Question

Two thin, isolated, concentric conducting spheres of radii 1.10 cm and 7.70 cm, have net charges of 13.50 nC and 4.00 nC, respectively. Use the sign of the field to indicate direction, thus a positive electric field would point away from the center of the spheres.
a) What is the electric field at 0.660 cm from the center of the spheres?
b) What is the electric potential at 0.660 cm from the center of the spheres?
c) What is the electric field at 3.740 cm from the center of the spheres?
d) What is the electric potential at 3.740 cm from the center of the spheres?
e) What is the electric field at 8.085 cm from the center of the spheres?
f) What is the electric potential at 8.09 cm from the center of the spheres?

I have found all of the electric fields but don't know how to find the electric potential.

Explanation / Answer

if a conducting sphere has a charge of Q and radius R,

then at any point inside this sphere , potential is given by

k*Q/R

where k=9*10^9

for any point outside the sphere, if distance from center is d,

then potential=k*Q/d

[ as you have calaculated the fields, i will calculate only potential values ]

part b:
0.66 cm is less that 1.1 cm and 7.7 cm

as the point is inside both the spheres,

potential calculation will involve radius of both the spheres.

potential=k*((13.5*10^(-9)/0.011)+(4*10^(-9)/0.077))=11512.98 volts

part d:
3.74 cm is outside sphere 1 but inside sphere 2

so for sphere 1, 3.74 cm distance will be used and for sphere 2, radius value will be used

then potential=k*((13.5*10^(-9)/0.0374)+(4*10^(-9)/0.077))=3716.2 volts

part f:

8.09 cm is outside both the spheres.

hence 8.09 cm distance will be used in the calculation for both the spheres

potential=k*((13.5*10^(-9)/0.0809)+(4*10^(-9)/0.0809))=1946.84 volts

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