Question

Electric field intensity is 400 V m⁻¹ at a distance of 2 m from a point charge. It will be 100 V m⁻¹ at a distance?
(a) 50 cm
(b) 4 cm
(c) 4 m
(d) 1.5 m

Answer 1

Electric Field Intensity of a point charge is given by:

$\boxed{E = \frac{1}{4\pi\varepsilon_{\circ}}\frac{q}{r^2}}$


Where


$q$ = Charge
$r$ = Distance from point charge.

We are given that the Electric Field Intensity is 400 V/m at a distance of 2 m. We have to find at what distance the Electric Field will be 100 V/m.

Let that distance be $r$.


Here we are not altering the charge. So we can say:

$E = \frac{1}{4\pi\varepsilon_{\circ}} \frac{q}{r^2} \\ \\ \\ \implies E \, r^2 = \frac{q}{4\pi\varepsilon_{\circ}} \\ \\ \\ \implies E \, r^2 = constant \\ \\ \\ \implies E_1 r_1 ^2 = E_2 r_2 ^2 \\ \\ \\ \implies 400 \times (2)^2 = 100 \times r^2 \\ \\ \\ \implies r^2 = \frac{400 \times 4}{100} \\ \\ \\ \implies r^2=16 \\ \\ \\ \implies \boxed{r=4 \, \, m}$


Thus, The Electric Field Intensity will be 100 V/m at a distance of 4 metres from the point charge.


So, the answer is Option (c) 4 m.

Answer 2

4m is the correct answer.