Question

A force of 2.25 N is acted on charge of 15 x 10-3 C.
Find the intensity of electric field of that point.​

Answer 1

Answer: The correct answer is 1500 N/C.

Explanation:

The expression for the electric field intensity is as follows;

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

q

F

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} Nq=15×10

−4

N .

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

15×10

−4

2.25

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

15×10

−4

2.25

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

Answer 2

Explanation:

Intensity of electric field =F/q

here F=2.25N

q=15×10-3c

therefore, Electric field=2.25/15×10-3

=2.25×1000/15=2250/15=150Nc-1