Chapter 21.2, Problem 33PP

To determine

The electric field strength so that the suspended drop is motionless.

Answer to Problem 33PP

  $4\times {10}^4\text{N/C}$ in downward direction.

Explanation of Solution

Given:

The weight of the drop, $W=6.4\times {10}^{-15}\text{N}$

The charge on the drop, $q=q_e=-1.6\times {10}^{-19}\text{C}$

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 drop 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 electric field must act in downward direction on the given negative charge.

  $\begin{array}{c}\left|F_e\right|=\left|F_g\right|\\ \left|qE\right|=W\\ E=\frac{6.4\times {10}^{-15}}{1.6\times {10}^{-19}}\\ E=4\times {10}^4\text{N/C}\end{array}$

Conclusion:

Hence, the electric field strength must be $4\times {10}^4\text{N/C}$ in downward direction.