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Q.

Electric potential at any point is V=−5x+3y+15z, then the magnitude of the electric field is

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a

32

b

42

c

52

d

7

answer is D.

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Detailed Solution

Ex=−dVdx=−(−5)=5; Ey=−dVdy=−3 and Ez=−dVdz=−15Enet =Ex2+Ey2+Ez2=(5)2+(−3)2+(−15)2=7
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