Question

A point charge gives rise to an electric field with magnitude 2 N/C at a distance of 4 m. If the distance is increased to 20 m, then what will be the new magnitude of the electric field?

Answer 1

Answer:

0.08 N/C

Explanation:

Electric Field: This is defined as the force per unit charge exerted at a point. The expression for electric field is given as,

E = Kq/r².............................. Equation 1

Where E = Electric Field, q = Charge, k = proportionality constant, r = distance.

making q the subject of the equation,

q = Er²/k............................... Equation 2

Given: E = 2 N/C, r = 4 m,

Substitute into equation 2

q = 2(4)²/k

q = 32/k C.

When r is increased to 20 m,

E = k(32/k)/20²

E = 32/400

E = 0.08 N/C.

Hence the electric Field = 0.08 N/C