Question

The magnitude of the electric field due to a point charge object at a distance of 4.0 m is 9 N/C. From the same charged object the electric field of magnitude, 16 N/C will be at a distance of ______.

पर्याय

• 1 m

• 2 m

• 3 m

• 6 m

Answer 1

The magnitude of the electric field due to a point charge object at a distance of 4.0 m is 9 N/C. From the same charged object the electric field of magnitude, 16 N/C will be at a distance of 3 m.

Explanation:

We know that,

E = `(kq)/r^2`

Where E is the magnitude of the electric field, k is constant, q is a charge of magnitude and r is the distance away from the point charge.

Given: E₁ = 9 N/C, r₁ = 4 and k = `9 xx 10^9` Nm²/C², E₂ = 16 N/C and q is same for both objects.

To find: r₂

`E_1 = (kq)/r_1^2`

`9 = ((9 xx 10^9 xx q))/4^2`

`q = ((9 xx 16))/((9 xx 10^9))`

q = 16 × 10^-9

`E_2 = (kq)/r_2^2`

`16 = ((9 xx 10^9 xx 16 xx 10^-9))/r_2^2`

`r_2^2 = 9`

`r_2 = 3` m