Question

a force of 2.25 N acts on a charge of 15×10-4C.The intensity of the electric field at that point is

Answer 1

Hope this helps you pal

Answer 2

Answer: The correct answer is 1500 N/C.

Explanation:

The expression for the electric field intensity is as follows;

$E=\frac{F}{q}$

Here, E is the electric field intensity, F is the force and q is the charge.

It is given in the problem that a force of 2.25 N acts on a charge of 15×10-4C.

Put F= 2.25 N and $q=15\times 10^{-4} N$.

$E=\frac{2.25}{15\times 10^{-4} }$

$E=\frac{2.25}{15\times 10^{-4} }$

Therefore, the intensity of the electric field is 1500 N/C.