Question

Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de Graaff).

Answer 1

Answer:

The magnitude of the electric field is 1.124 X 10⁷ N/C

Explanation:

Magnitude of electric field is given as;

$E = \frac{kq}{r^2} , N/C$

where;

E is the magnitude of the electric field, N/C

q is the point charge, C

k is coulomb's constant, Nm²/C²

r is the distance of the point charge, m

Given;

q = 5mC = 5×10⁻³ C

r = 2m

k = 8.99 × 10⁹ Nm²/C²

Substitute these values and solve for magnitude of electric field E

$E = \frac{kq}{r^2} = \frac{(8.99 X10^9)(5X10^{-3})}{2^2} = 1.124 X10^7 N/C$

Therefore, the magnitude of the electric field is 1.124 X 10⁷ N/C