Answer to Question #183073 in Electricity and Magnetism for KyKy

Question

A proton is released from rest in a uniform horizontal electric field. It travels 3.25 m for 5 μs. Find the acceleration of the proton and the magnitude of the electric field.

Answer 1

To be given in question distance (s)=3.25meter

t=5$\mu sec$

M$=m_{p}$

$q_{p}=q_{e}$

$u=0$

To be asked in question

acceleration (a)=?

Electric force ( F).=?

We know that

Newton's motion 2^nd law

$S=ut+\frac{1}{2}at^2$

$3.25=0+\frac{1}{2}a\times(5\times10^{-6})^2$

$a=$ 2.6$\times10^{11}meter/sec^2$

Electric force

We know that

$qE =ma$

$E=\frac{m_{p}a}{q_{p}}$

Put value

$E =\frac{1.67\times10^{-27}\times2.6\times10^{11}}{1.6\times10^{-19}}$

$E=2.71375 \times10^{3}N/C$