Question

A 5.0g conducting sphere with charge of 20 μC hangs by a non-conducting thread in an electric field produced by two plates separated by 8.0cm. What potential will cause the ball to hang at 25° to the vertical? 

Answer 1

Let’s write the conditions of the equilibrium for the conducting sphere:

$$\sum F_x=0, \sum F_y=0$$

Let’s consider the forces that act on the conducting sphere in the horizontal $x$- and vertical $y$-direction:

$$Tsin\theta-F_e=0, (1)$$ $$Tcos\theta-mg=0. (2)$$

We can express the force $T$ from the second equation:

$$T=\dfrac{mg}{cos\theta}.$$

Then we can substitute it into the first equation and find the electrostatic force, $F_e$:

$$mgtan\theta-F_e=0,$$ $$F_e=mgtan\theta.$$

From the other hand, $F_e=qE.$ From this formula we can find the electic field strength:

$$E=\dfrac{F_e}{q}=\dfrac{mgtan\theta}{q}.$$

Finally, we can find the potential that will cause the ball to hang at 25° to the vertical from the formula:

$$\Delta V=Ed=\dfrac{mgdtan\theta}{q},$$ $$\Delta V=\dfrac{5.0\cdot10^{-3}\ kg\cdot 9.8\ \dfrac{m}{s^2}\cdot 8\cdot10^{-2}\ m\cdot tan25^{\circ}}{2\cdot10^{-5}\ C}=91.4\ V$$

Answer:

$\Delta V=91.4\ V$