Question

A charged particle of mass $4\times {10}^{-13}$ kg is hanging stationary between two horizontal charged plates separated by $2$ cm and potential difference between plates is $3.2\times {10}^4$ V. Find the charge of the particle ?

• A. $2.8\times {10}^{-15}C$
• B. $2.4\times {10}^{-20}C$
• C. $2.4\times {10}^{-16}C$
• D. $2.5\times {10}^{-18}C$

Answer 1

The correct option is D $2.5\times {10}^{-18}C$

As we know

$E=\frac{V}{d}=\frac{3.2\times {10}^4}{2\times {10}^{-2}}=1.6\times {10}^6$

Now as we know $F=qE=>mg=q\times 1.6\times {10}^6$

$⟹4\times {10}^{-13}\times 10=q\times 1.6\times {10}^6$

$q=2.5\times {10}^{-18}C$