A uniform electric field of 400 V/m is directed at 45∘ above the x-axis as shown in the figure. The potential difference VA−VB is given by :
0
4 V
6.4 V
2.8 V
A
0
B
4 V
C
6.4 V
D
2.8 V
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Solution
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Here, the electric field, →E=Ecos45^i+Esin45^j=E√2(^i+^j) now, VA−VB=∫BA→E.→dr=∫BAE√2(^i+^j).(dx^i+dy^j)=E√2[∫0.030dx+∫00.02dy] or VA−VB=400√2[0.03−0.02]=2.8V
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