Question

What is the strength of the electric field ep 0.90 mm from a proton?

Answer 1

The electric field strength is inversely related to the square of the distance.so  the strength of the electric field is given by

$E=\frac{1}{4\pi \epsilon _{0} } \frac{q}{r^{2} } = \frac{k q}{r^{2} }$

Here,  $\frac{1}{4\pi \epsilon _{0} } = k$  is constant depend upon medium and its value is $9.0 \times10^{9} \ N m^2/C^2$ and q is charge and  r is the distance.

Given  $r = 0.90 mm = 9.0 \times 10^{-4} m$ and we know the charge of proton, $q = 1.6\times 10^{-19} \ C$.

Therefore,

$E=\frac{9.0 \times10^{9} \ N m^2/C^2 \times1.6\times 10^{-19} \ C }{(9.0 \times 10^{-4} m)^2} = 0.177 \times 10^{-2} \ N/C \\\\ E= 1.77 \times 10^{-3} N/C$