Question

Two 5.0-cm-diameter conducting spheres are 8.0 m apart, and each carries 0.12 mC. Determine (a) the potential on each sphere, (b) the field strength at the surface of each sphere, (c) the potential midway between the spheres, and (d) the potential difference between the spheres. Wolfson, Richard. Essential University Physics, Volume 2 (p. 432). Pearson Education. Kindle Edition.

Answer 1

Answer:

Explanation:

Two spheres 10m apart

Each charge on the sphere is 0.12mC= 0.12×10^-3C

Given that the diameter is

5cm=0.05m

Then, the radius is diameter / 2

r=d/2

r=0.05/2

r=0.025m

Potential is given as

V=kq/r

k=9×10^9Nm2/C2

a. Potential on each sphere surface.

They are going to have the same value since the sphere are identical

At the surface of the sphere,

r= 0.025m

V=kq/r

V=9E9×0.12E-3/0.025

V=4.32E7Volts

V=4.32×10^7Volts

b. The electric field at the surface of each sphere will be the same since the charge are identical,

So, Electric field is given as

E=kq/r^2

At the surface

E=9E9×0.12E-3/0.025^2

E=1.728×10^8N/C

c. The potential mid way between the two sphere

The potential difference due to the first sphere

The sphere are 10m apart then the distance mid way is 5m

Then, the radius of the sphere is 0.025m

Total distance from the center is 5.025m

Then,

V=kq/r

V1=9E9×0.12E-3/5.025

V1=2.149×10^5 Volts

Potential difference due to the second sphere is the same as the first, since both are identical

Total potential is V1 +V2

V=2.149E5+2.149E5

V=4.3E5Volts

Total potential at the middle due to the two sphere is 4.3×10^5Volts

d. The potential difference between the sphere at any point is equal to the potential difference found at c

Therefore the potential difference is

V=4.3×10^5Volts