A charge of 5 C experiences a force of 5000 N when it is kept in a uniform electric field. What is the potential difference between two points separated by a distance of 1 cm ?

Question

A charge of 5 C experiences a force of 5000 N when it is kept in a uniform electric field. What is the potential difference between two points separated by a distance of 1 cm ? (A) 10 V (B) 250 V (C) 1000 V (D) 2500 V

Tardigrade Admin · Accepted Answer

F=Q E=(Q V/d) ⇒ 5000 =(5 × V/10-2) ⇒ V=10 volt

Q. A charge of $5 C$ experiences a force of $5000 N$ when it is kept in a uniform electric field. What is the potential difference between two points separated by a distance of $1 c m ?$

10 V

250 V

1000 V

2500 V

$F = QE = \frac{Q V}{d}$
$\Rightarrow 5000 = \frac{5 \times V}{1 0 ^{- 2}}$
$\Rightarrow V = 10$ volt

Q. A charge of 5C experiences a force of 5000N when it is kept in a uniform electric field. What is the potential difference between two points separated by a distance of 1cm?