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

Question

A force of 2.25 N acts on a charge of 15 × 10-4 C. The intensity of electric field at that point is (A) 150 NC-1 (B) 15 NC-1 (C) 1500 NC-1 (D) 1.5 NC-1

Tardigrade Admin · Accepted Answer

Electric field, E=(F/q) =(2.25 N/15×10-4 C) =1500 N C-1

Q. A force of $2.25 N$ acts on a charge of $15 \times 1 0^{- 4} C$ . The intensity of electric field at that point is

$150 N C^{- 1}$

$15 N C^{- 1}$

$1500 N C^{- 1}$

$1.5 N C^{- 1}$

Electric field, $E = \frac{F}{q}$
$= \frac{2.25 N}{15 \times 1 0 ^{- 4} C}$
$= 1500 N C^{- 1}$

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

150NC−1

15NC−1

1500NC−1

1.5NC−1