Question

A proton is shot perpendicularly at an infinite plane of charge. The charge density of the plane is +7.65×10^−4 C/m^2. If the proton starts out 1.05 mm from the plane, what must its velocity be so that it screeches to a halt exactly as it reaches the plane?

Answer 1

Answer:

The velocity is $2.94\times10^{6}\ m/s$.

Explanation:

Given that,

Charge density of the plane $\sigma=7.65\times10^{−4}\ C/m^2$

Distance = 1.05 mm

We need to calculate the electric field due to plane of charge

Using formula of electric field

$E=\dfrac{\sigma}{2\epsilon}$

Put the value into the formula

$E=\dfrac{7.65\times10^{−4}}{2\times8.85\times10^{-12}}$

$E=4.322\times10^{7}\ N/C$

We need to calculate potential difference

Using formula of potential difference

$V=E\times r$

Put the value into the formula

$V=4.322\times10^{7}\times1.05\times10^{-3}$

$V=4.5381\times10^{4}\ Volt$

We need to calculate the work requires to be done to reach the surface of the plane

Using formula of work done

$W=qV$

Put the value into the formula

$W = 1.6\times10^{-19}\times4.5381\times10^{4}$

$W=7.26096\times10^{-15}\ J$

We need to calculate the velocity

Using work energy theorem

$W=\dfrac{1}{2}mv^2$

$v^2=\dfrac{2W}{m}$

$v=\sqrt{\dfrac{2\times7.26096\times10^{-15}}{1.67\times10^{-27}}}$

$v=2.94\times10^{6}\ m/s$

Hence, The velocity is $2.94\times10^{6}\ m/s$.