Electrostatics Question 442

Question: Electric potential at any point is V=−5x+3y+15z , then the magnitude of the electric field is [MP PET 2002]

Options:

A) 32

B) 42

C) 52

D) 7

Show Answer

Answer:

Correct Answer: D

Solution:

Ex=−dVdx=−(−5)=5;

Ey=−dVdy=−3 and Ez=−dVdz=−15

Enet=Ex2+Ey2+Ez2=(5)2+(−3)2+(−15)2=7

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