Two metal spheres, each of radius 2.0 cm, have a center-to-center separation of 2.7 m. One has a charge of +1.0 ✕ 10-8 C; the other has a charge of -3.7 ✕ 10-8 C. Assume that the separation is large enough relative to the size of the spheres to permit us to consider the charge on each to be uniformly distributed (the spheres are electrically isolated from each other) Take V = 0 at infinity.
A) Calculate the potential at the point halfway between the centers.
B) Calculate the potential on the surface of sphere 1.
C) Calculate the potential on the surface of sphere 2.

Explanation:

potential half way

= k q / r

9 x 10⁹ x 10⁻⁸ x 2 / 2.7

= 66.67 V

potential due to second charge

k q / r

- 9 x 10⁹ x 10⁻⁸ x 3.7 X2  / 2.7

= -246.67 V

Potential at half way point

= 66.67 - 246.67

= - 180 V  

B) Potential at the surface of sphere 1

= kq / r

9 x 10⁹ x 10⁻⁸ / .02

= 4500 V

We have neglected the effect of other charge due to comparatively large distance

Potential at the surface of sphere 2

= kq / r

=  - 9 x 10⁹ x3.7x 10⁻⁸ / .02

= - 16650 V

We have neglected the effect of other charge due to comparatively large distance

change tension

the answer is 100. i just did this lesson and got a 100% on it. : )

-fishtank