- #1
binbagsss
- 1,269
- 11
Given the standard formulas of an insulating sphere, inside and outside respectively:
[1] Q/4πΣσr^2
[2] Qr/4πΣσR^3
- where R referes to the radius of the solid insulating sphere.
Charged paint is spread in a very thin uniform layer over the surface of a plastic sphere o diameter 12.0cm, gicing it a charge of -35*10^-6C, Find E:
a) just outside the paint layer?
b) just inside the paint layer?
c) 5.00cm outside the surface of the paint layer?
My Attempt:
a) just outside: Using [1] with r=6.0cm and Q = -35*10^-6 , I obtain 8.74*10^7N/C , is the correct answer
b) just inside: E = 0 (From Gauss's Law)
c) just outside: Using [1] with r = 5.0cm and R = 6.0cm, I obtain 1.26*10^8N/C, however the correct answer is 2.60*10^7N/C
If anyone could point my in the right direction for c, greatly appreciated, ta =]
[1] Q/4πΣσr^2
[2] Qr/4πΣσR^3
- where R referes to the radius of the solid insulating sphere.
Charged paint is spread in a very thin uniform layer over the surface of a plastic sphere o diameter 12.0cm, gicing it a charge of -35*10^-6C, Find E:
a) just outside the paint layer?
b) just inside the paint layer?
c) 5.00cm outside the surface of the paint layer?
My Attempt:
a) just outside: Using [1] with r=6.0cm and Q = -35*10^-6 , I obtain 8.74*10^7N/C , is the correct answer
b) just inside: E = 0 (From Gauss's Law)
c) just outside: Using [1] with r = 5.0cm and R = 6.0cm, I obtain 1.26*10^8N/C, however the correct answer is 2.60*10^7N/C
If anyone could point my in the right direction for c, greatly appreciated, ta =]