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Question: Two metal pieces having a potential difference of \(800V\) are \(0.02m\) apart horizontally. A parti...

Two metal pieces having a potential difference of 800V800V are 0.02m0.02m apart horizontally. A particle of mass 1.96×10−15kg1.96 \times 10^{- 15}kgis suspended in equilibrium between the plates. If ee is the elementary charge, then charge on the particle is

A

ee

B

3e3e

C

6e6e

D

8e8e

Answer

3e3e

Explanation

Solution

For equilibrium mg = qE

∴ 1.96×10−15×9.8=q×(8000.02)1.96 \times 10^{- 15} \times 9.8 = q \times \left( \frac{800}{0.02} \right)

⇒ q=1.96×10−15×9.8×0.02800q = \frac{1.96 \times 10^{- 15} \times 9.8 \times 0.02}{800}

⇒ n×1.6×10−19=1.96×10−15×9.8×0.02800n \times 1.6 \times 10^{- 19} = \frac{1.96 \times 10^{- 15} \times 9.8 \times 0.02}{800}⇒ n = 3.