An electric charge 10-3-mu C is placed the origin -0-0- of X - Y co-ordinate system- Two points A and B are situated -sqrt-2-sqrt-2- and -2- 0- respectively- The potential difference between the points A and B will be-9 voltzero2 volt4-5 volt

Question

Toppr Admin · Accepted Answer

-VA-14-x3C0-x3F5-qrA-14-x3C0-x3F5-q-x221A-2-2-14-x3C0-x3F5-q2-VB-14-x3C0-x3F5-qrB-14-x3C0-x3F5-q-x221A-4-0-14-x3C0-x3F5-q2-VA-x2212-VB-0-

An electric charge $10^{- 3} μ C$ is placed at the origin $(0, 0)$ of X - Y co-ordinate system. Two points A and B are situated at $(\sqrt{2}, \sqrt{2})$ and $(2, 0)$ respectively. The potential difference between the points A and B will be:
$9$ volt
zero
$2$ volt
$4.5$ volt

$V_{A} = \frac{1}{4 π ϵ} \frac{q}{r_{A}} = \frac{1}{4 π ϵ} \frac{q}{\sqrt{2 + 2}} = \frac{1}{4 π ϵ} \frac{q}{2}$

$V_{B} = \frac{1}{4 π ϵ} \frac{q}{r_{B}} = \frac{1}{4 π ϵ} \frac{q}{\sqrt{4 + 0}} = \frac{1}{4 π ϵ} \frac{q}{2}$

$V_{A} - V_{B} = 0$

An electric charge 10−3μC is placed at the origin (0,0) of X - Y co-ordinate system. Two points A and B are situated at (√2,√2) and (2,0) respectively. The potential difference between the points A and B will be:9 voltzero2 volt4.5 volt

VA=14πϵqrA=14πϵq√2+2=14πϵq2 VB=14πϵqrB=14πϵq√4+0=14πϵq2 VA−VB=0

An electric charge $10^{- 3} μ C$ is placed at the origin $(0, 0)$ of X - Y co-ordinate system. Two points A and B are situated at $(\sqrt{2}, \sqrt{2})$ and $(2, 0)$ respectively. The potential difference between the points A and B will be:
$9$ volt
zero
$2$ volt
$4.5$ volt

$V_{A} = \frac{1}{4 π ϵ} \frac{q}{r_{A}} = \frac{1}{4 π ϵ} \frac{q}{\sqrt{2 + 2}} = \frac{1}{4 π ϵ} \frac{q}{2}$

$V_{B} = \frac{1}{4 π ϵ} \frac{q}{r_{B}} = \frac{1}{4 π ϵ} \frac{q}{\sqrt{4 + 0}} = \frac{1}{4 π ϵ} \frac{q}{2}$

$V_{A} - V_{B} = 0$