The potential difference between to points A and B VA−VB in an electric E¯=2i+3j+4k¯NC−1 where rA¯=i−2j+K¯m and rB¯=2i+j−2K¯m is

VA−VB=−E→.dr→dr→=rB→-rA→=2i^+j^+-2k^-i^+3j^-3k^=i^+3 j^-3k^-E→.dr→=2i^+3j^+4k^.i^+3 j^-3k^=-1V Solving VA−VB=−1V

The potential difference between to points A and B VA−VB in an electric E¯=2i+3j+4k¯NC−1 where rA¯=i−2j+K¯m and rB¯=2i+j−2K¯m is

VA−VB=−E→.dr→dr→=rB→-rA→=2i^+j^+-2k^-i^+3j^-3k^=i^+3 j^-3k^-E→.dr→=2i^+3j^+4k^.i^+3 j^-3k^=-1V Solving VA−VB=−1V

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The potential difference between to points A and B $(V_{A} - V_{B})$ in an electric $\bar{E} = (2 i + 3 j + 4 \bar{k}) N C^{- 1}$ where $\bar{r_{A}} = (i - 2 j + \bar{K}) m$ and $\bar{r_{B}} = (2 i + j - 2 \bar{K}) m$ is

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-2V

-1V

-4V

-6V

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Detailed Solution

$V_{A} - V_{B} = - (\vec{E} . \vec{d r})$

$\vec{d r} = \vec{r_{B}} - \vec{r_{A}} = (2 \hat{i} + \hat{j} + - 2 \hat{k}) - (\hat{i} + 3 \hat{j} - 3 \hat{k}) = \hat{i} + 3 \hat{j} - 3 \hat{k}$

$- \vec{E} . \vec{d r} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) . (\hat{i} + 3 \hat{j} - 3 \hat{k}) = - 1 V$

$Solving V_{A} - V_{B} = - 1 V$

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The potential difference between to points A and B VA−VB in an electric E¯=2i+3j+4k¯NC−1 where rA¯=i−2j+K¯m and rB¯=2i+j−2K¯m is

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The potential difference between to points A and B $(V_{A} - V_{B})$ in an electric $\bar{E} = (2 i + 3 j + 4 \bar{k}) N C^{- 1}$ where $\bar{r_{A}} = (i - 2 j + \bar{K}) m$ and $\bar{r_{B}} = (2 i + j - 2 \bar{K}) m$ is