Chapter 21.1, Problem 3PP

To determine

The net charge(magnitude and sign) to be placed on the pith ball so that electrostatic ball acting on it is balanced by the gravitational force.

Answer to Problem 3PP

  $-0.323\times {10}^{-7}\text{C}$

Explanation of Solution

Given:

The weight of the pith ball, $W=2.1\times {10}^{-3}\text{N}$

The electric field acting on the ball in downward direction, $E=6.5\times {10}^4\text{N/C}$

Formula used:

The electric field strength is given by,

  $E=\frac{F_{\text{on }q\text{'}}}{q\text{'}}$

Where,

  $F_{\text{on }q\text{'}}=\text{Electric force on }q\text{'}$

Calculation:

It is known that the positive charge moves in the direction of Electric field while negative charge moves in opposite direction of Electric field.

Now, the gravitational force on the ball will act in vertically downward direction, magnitude of which is equal to the weight of the ball. Thus, the electrostatic force must act in the upward direction to balance it by the gravitational force. Therefore, the sign of the charge must be negative since the electric field is in vertical downward direction and the magnitude of the charge must be,

  $\begin{array}{l}\left|F_e\right|=\left|F_g\right|\\ \left|qE\right|=W\\ q=\frac{2.1\times {10}^{-3}}{6.5\times {10}^4}\\ q=0.323\times {10}^{-7}\text{C}\end{array}$

Conclusion:

Hence, the charge of the pith ball must be $-0.323\times {10}^{-7}\text{C}$ .