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Find potential difference $\;V_{AB}\;$ between $\;A(0,0,0)$ and $\;B(1m ,1m, 0)\;$ in an electric field $\;\overrightarrow{E}=y \hat{i}+x \hat{j}$

$(a)\;0 V\qquad(b)\;1 V\qquad(c)\;2 V\qquad(d)\;None$

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Answer : (b) 1V
Explanation :
$dV= - \overrightarrow{E}\;.\overrightarrow{dr}$
$\int_{B}^{A}\;dV = -\int_{(1, 1, 0)}^{(0 ,0 ,0)}\;(y \hat{i} +x \hat{j}) .(dx \hat{i}+dy \hat{j} +dz \hat{k})$
$V_{A}-V_{B}= - \int_{(1 ,1 .0)}^{(0 ,0,0)}\;(y \;dx +x \;dy)$
$V_{AB}=-\int_{(1 ,1 , 0)}^{(0,0,0)}\;d(xy)$
$V_{AB}=- [xy]_{(1,1,0)}^{(0,0,0)}$
$V_{AB}=1 V$
answered Feb 7, 2014 by yamini.v
 

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