Answer to Question #142411 in Physics for Muhammad Kamran

Question

A metal ball of radius 12 cm is charged to potential 200 V. A point charge 3 nC is moved from infinity to a point near the ball surface. In doing so, work 100 nJ is performed. Find the distance from the final position of the charge to the ball surface. The answer should be expressed in centimeters [cm]. The ball is in a vacuum.

Answer 1

As we know, the work is

$$W=\Delta\phi q_2.$$

The potential difference $\Delta\phi=(200-0)=200\text{ V},$ charge 2 is 3 nC.

The potential of the ball with radius R is

$$\phi_1=\frac{kq_1}{R}.$$

Thus:

$$q_1=\frac{R\phi_1}{k}.$$

At any distance greater than R the ball creates a potential as of a point charge. The work can be expressed as

$$W=\frac{kq_1q_2}{r},\\\space\\ r=\frac{kq_1q_2}{W}=\frac{\phi_1q_2R}{W}=0.72\text{ m}.$$

Thus, the distance to the surface of the ball is 60 m.