In: Physics
The electric potential immediately outside a charged conducting sphere is 220 V, and 10.0 cm farther from the center of the sphere the potential is 140 V.
(a) Determine the radius of the sphere.
. cm
(b) Determine the charge on the sphere.
nC
The electric potential immediately outside another charged conducting sphere is 250 V, and 10.0 cm farther from the center the magnitude of the electric field is 360 V/m. (c) Determine all possible values for the radius of the sphere. (Enter your answers from smallest to largest. If only one value exists, enter "NONE" in the second answer blank.)
r1 = | . cm |
r2 = | . cm |
(d) Determine the charge on the sphere for each value of
r. (If only one value exists, enter "NONE" in the second
answer blank.)
q1 = | . nC |
q2 = | nC |
Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.
The electric potential immediately outside a charged conducting
sphere is 240 V, and 10.0 cm farther from the center of the sphere
the potential is 130 V.
(a) Determine the radius of the sphere.
____ cm
(b) Determine the charge on the sphere.
____ nC
The electric potential immediately outside another charged
conducting sphere is 270 V, and 10.0 cm farther from the center the
magnitude of the electric field is 410 V/m.
(c) Determine all possible values for the radius of the sphere.
(Enter your answers from smallest to largest. If only one value
exists, enter "NONE" in the second answer blank.)
r1 = cm
r2 = cm
(d) Determine the charge on the sphere for each value of r. (If
only one value exists, enter "NONE" in the second answer
blank.)
q1 = nC
q2 = nC
If the radius is r metres, then going from r to (r + 0.1) causes
V to drop from 180V to 160V.
180 = kq/r
160 = kq/(r+0.1)
Dividing gives:
180/160 = (r+0.1)/r
1.125 = 1 + 0.1/r
0.1/r= 0.125
r= 0.1/0.125=0.8m
q = 180r/k = 180 x 0.8 /9x10^9 = 1.6x10^-8 C