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In a uniform electric field , the potential is $10\;V$ at the origin of coordinates and $8V$ at each of the points $(1 ,0 ,0) ,(0 , 1 ,0)\;and\;(0 ,0,1)\;.$ The potential at the point $\;(1 ,1 ,1)\;$ will be

$(a)\;0\qquad(b)\;4V\qquad(c)\;8V\qquad(d)\;10V$

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Explanation :
Let the electric field vector $\;\overrightarrow{E}=E_{x} \hat{i} +E_{y} \hat{j} +E_{z} \hat{k}$
$V_{A}-V_{B}=-\int_{B}^{A}\;\overrightarrow{E} . dr$
$V_{(1 , 0 , 0)}-V_{(0 , 0 ,0)}=-\int_{(0 , 0 ,0)}^{(1 ,0 ,0)}\;E_{x}=[-E_{x}\;x]_{(0 , 0 ,0)}^{(1 , 0 ,0)}$
$8-10=-E_{x}$
$E_{x}=2$
Similarly $\;V_{(0,1,0)}-V_{(0 , 0, 0)}=-\int_{(0 , 0 ,0)}^{(0 , 1 ,0)}\;E_{y} dy$
$-2=-E_{y}\times1$
Similarly $E_{z}=2$
Thus
$\overrightarrow{E}=2 \hat{i} + 2 \hat {j} + 2 \hat{k}$
Now
$V_{(1 ,1 ,1)}-V_{(0 , 0 ,0)}=-\int _{(0 , 0 ,0)}^{(1 ,1,1)}\;(2 \hat{i}+2 \hat{j}+ 2 \hat{k})\;.(d_{x} \hat{i} +d_{y} \hat{j} +d_{z} \hat{k})$
$=-\int_{(0 ,0 ,0)}^{(1 , 1 .1)}\;{(2 dx + 2 dy + 2 dz)}$
$=-(2\times1+2\times1+2\times1)$
$V_{(1 ,1 , 1)}-10=-6$
$V_{(1 ,1 ,1)}=4V\;.$

 

answered Feb 18, 2014 by yamini.v
edited Mar 23, 2014 by meenakshi.p
 

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