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# A uniform electric field of 100 V/m is directed at $30^{0}$ with the +ve x - axis as shown in figure . Find the potential difference $\;V_{BA}\;$ if OA=2 m and OB= 4 m

$(a)\;100\;(2+\sqrt{3})\qquad(b)\;-100\;(2+\sqrt{3})\qquad(c)\;100\;\sqrt{3}\qquad(d)\;-200$

Answer : (c) $\;100\;\sqrt{3}$
Explanation : Electric field in vector form can be written as ,
$\overrightarrow{E}=(100\;cos\;30^{0}\;\hat{i}+100\;sin\;30^{0}\;\hat{j})$
$\overrightarrow{E}=50\;\sqrt{3}\;\hat{i}+50\;\hat{j}$
$A=(-2m , 0)$
$B=(0 , 4m)$
$V_{BA}=V_{B}-V_{A}=-\int_{A}^{B}\;\overrightarrow{E}\;.\overrightarrow{dr}$
$=-\int_{(-2 ,0)}^{(0,4)}\;(50\;\sqrt{3}\;dx+50\;dy)$
$=-\;[50\;\sqrt{3}\;x+50\;y]_{(-2 ,0)}^{(0 , 4)}$
$=-(100\;\sqrt{3}+200)$
$=-100\;(2+\sqrt{3})\;.$