First slide

Electrostatic potential

Question

Two metal pieces having a potential difference of 800 V are 0.02 m apart horizontally. A particle of mass $1 .96 \times 10^{- 15} kg$ is suspended in equilibrium between the plates. If e is the elementary charge, then charge on the particle is

Easy

A

15e
B

3e
C

6e
D

8e

Solution

For equilibrium mg = qE
$∴ 1 .96 \times 10^{- 15} \times 9 .8 = q \times (\frac{800}{0 .02})$
$\Rightarrow q = \frac{1 .96 \times 10^{- 15} \times 9 .8 \times 0 .02}{800}$
$\Rightarrow n \times 1 .6 \times 10^{- 19} = \frac{1 .96 \times 10^{- 15} \times 9 .8 \times 0 .02}{800} \Rightarrow n = 3$

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