A point charge of 2 C experiences a constant force of 1000 N when moved between two points separated by a distance of 2 cm in a uniform electric field. The potential difference between the two points is

Question

A point charge of 2 C experiences a constant force of 1000 N when moved between two points separated by a distance of 2 cm in a uniform electric field. The potential difference between the two points is (A) 12 V (B) 8 V (C) 10 V (D) 16 V

Tardigrade Admin · Accepted Answer

Given, q=2 C , F=100 N and d=2 × 10-2 m Electric field E=(F/q) =(1000/2)=500 N / C The potential difference V =E × d =500 × 2 × 10-2 =103 × 10-2 =10 V

Q. A point charge of $2 C$ experiences a constant force of $1000 N$ when moved between two points separated by a distance of $2 c m$ in a uniform electric field. The potential difference between the two points is

12 V

8 V

10 V

16 V

5 V

Given,
$q = 2 C, F = 100 N$ and $d = 2 \times 1 0^{- 2} m$
Electric field $E = \frac{F}{q}$
$= \frac{1000}{2} = 500 N / C$
The potential difference
$V = E \times d$
$= 500 \times 2 \times 1 0^{- 2}$
$= 1 0^{3} \times 1 0^{- 2}$
$= 10 V$

Q. A point charge of 2C experiences a constant force of 1000N when moved between two points separated by a distance of 2cm in a uniform electric field. The potential difference between the two points is

12 V

8 V

10 V

16 V

5 V

Given, q=2C,F=100N and d=2×10−2m Electric field E=qF​ =21000​=500N/C The potential difference V=E×d =500×2×10−2 =103×10−2 =10V

Q. A point charge of $2 C$ experiences a constant force of $1000 N$ when moved between two points separated by a distance of $2 c m$ in a uniform electric field. The potential difference between the two points is

Given,
$q = 2 C, F = 100 N$ and $d = 2 \times 1 0^{- 2} m$
Electric field $E = \frac{F}{q}$
$= \frac{1000}{2} = 500 N / C$
The potential difference
$V = E \times d$
$= 500 \times 2 \times 1 0^{- 2}$
$= 1 0^{3} \times 1 0^{- 2}$
$= 10 V$