An electric charge 10^{-3} mu C is placed at | vedclass

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Question

The distance of point $A(\sqrt{2}, \sqrt{2})$ from the origin,

$OA=\left|\vec{r}_{1}\right|=\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}=\sqrt{4}=2$ units.

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Vedclass · Answer

The distance of point $A(\sqrt{2}, \sqrt{2})$ from the origin,

$OA=\left|\vec{r}_{1}ight|=\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}=\sqrt{4}=2$ units.

The distance of point $B(2,0)$ from the origin,

$O B=|\overrightarrow{r_{2}}|=\sqrt{(2)^{2}+(0)^{2}}=2$ units.

Now, potential at $A, V_{A}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{Q}{(O A)}$

Potential at $B, V_{B}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{Q}{(O B)}$

$	herefore $ Potential difference between the points $A$ and $B$ is zero.

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

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