Two points A and B are located in diametrically opposite directions of a point charge of +2 μ C at distances 2 m and 1 m respectively from it. The potential difference between A and B is

Question

Two points A and B are located in diametrically opposite directions of a point charge of +2 μ C at distances 2 m and 1 m respectively from it. The potential difference between A and B is (A) 3 × 103 V (B) 6 × 104 V (C) -9 × 103 V (D) -3 × 103 V

Tardigrade Admin · Accepted Answer

Here, q=2 μ C=2×10-6 C, rA=2m, rB=1 m ∴ VA-VB=(q/4πε0) [(1/rA)-(1/rB)] =2 ×10-6×9×109[(1/2)-(1/1)]V =-9 ×103 V

Q. Two points $A$ and $B$ are located in diametrically opposite directions of a point charge of $+ 2 μ C$ at distances $2 m$ and $1 m$ respectively from it. The potential difference between $A$ and $B$ is

$3 \times 1 0^{3} V$

$6 \times 1 0^{4} V$

$- 9 \times 1 0^{3} V$

$- 3 \times 1 0^{3} V$

Here, $q = 2 μ C = 2 \times 1 0^{- 6} C$ ,
$r_{A} = 2 m, r_{B} = 1 m$
$∴ V_{A} - V_{B} = \frac{q}{4 π ε _{0}} [\frac{1}{r _{A}} - \frac{1}{r _{B}}]$
$= 2 \times 1 0^{- 6} \times 9 \times 1 0^{9} [\frac{1}{2} - \frac{1}{1}] V$
$= - 9 \times 1 0^{3} V$

Q. Two points A and B are located in diametrically opposite directions of a point charge of +2μC at distances 2m and 1m respectively from it. The potential difference between A and B is

3×103V

6×104V

−9×103V

−3×103V