Question

A proton is released from rest, 10 cm from a charged sheet carrying charged density of $-2.21\times {10}^{-9}C/m^2$. It will strike the sheet after the time (approximately)

• A. $4\mu S$
• B. $2\mu s$
• C. $2\sqrt{2}\mu S$
• D. $4\sqrt{2}\mu s$

Answer 1

The correct option is A $4\mu S$

Mass of proton $m=1.67\times {10}^{-27}$ kg and charge $=e=1.6\times {10}^{-19}C$

Magnitude of electric field due to charge sheet is $E=\frac{\sigma }{2{ϵ}_0}=\frac{2.21\times {10}^{-9}}{2\times 8.85\times {10}^{-12}}=124.86N/C$

Force on proton due to charge sheet is $F=eE=1.6\times {10}^{-19}\times 124.86$

or $ma=F=1.6\times {10}^{-19}\times 124.86$

or $a=\frac{1.6\times {10}^{-19}\times 124.86}{1.67\times {10}^{-27}}=1.2\times {10}^{10}m/s^2$

Using, $S=ut+\frac{1}{2}a{t}^2,$

here, $u=0$ so, $S=\frac{1}{2}a{t}^2$

or $t=\sqrt{\frac{2S}{a}}=\sqrt{\frac{2\times 0.1}{1.2\times {10}^{10}}}=4.08\times {10}^{-6}sec\sim 4\mu$ Sec