Question

Two horizontal metal plates having a potential difference of $800V$ are $0.02m$ apart. A particle of mass $1.92\times {10}^{-15}kg$ is suspended in equilibrium between the plates. Ife is the elementry charge, then charge on the particle is:

• A. $e$
• B. $3e$
• C. $6e$
• D. $8e$

Answer 1

The correct option is B $3e$

$\begin{array}{c}V=800V\\ d=0.02m\\ m=1.92\times {10}^{-5}kg\end{array}$

For equilibrium of particle

weight of particle = Electrostatic force

$\begin{array}{c}mg=gE\\ mg=q\frac{V}{d}\\ q=\frac{mgd}{V}\\ =\frac{1.92\times {10}^{-5}\times 9.8\times 0.02}{800}\\ =4.8\times {10}^{-19}C\end{array}$

If e is the elementary charge then

$\begin{array}{c}q=3eC\\ wheree=1.6\times {10}^{-19}coulomb\end{array}$

Option B.