Question

There is a -5 nC charge present.A. Determine the magnitude and direction of the electric field at a point 7.3 meters from this charge.B. How far from this charge does the electric field have a magnitude of 2 N/C?

Answer 1

Answer:

(A) E = 0.84 N/C

(B) r = 4.74 meters

Explanation:

Given that,

Charge, $q=-5\ nC=-5\times 10^{-9}\ C$

(a) The magnitude of electric field at a point 7.3 meters from this charge is given by :

$E=\dfrac{kq}{r^2}$

$E=\dfrac{9\times 10^9\times 5\times 10^{-9}}{(7.3)^2}$

E = 0.84 N/C

We know that electric field due to negative charge is inwards. So, the direction of electric field is towards the charge.

(b) Let at a distance of d meters the electric field have a magnitude of 2 N/C. It is given by :

$E=\dfrac{kq}{r^2}$

$r=\sqrt{\dfrac{kq}{E}}$

$r=\sqrt{\dfrac{9\times 10^9\times 5\times 10^{-9}}{2}}$

r = 4.74 meters

Hence, this is the required solution.