Potential Gradient and Relation between Electric Field and Potential
Electric pote...
Question
Electric potential is given by V=6xā8xy2ā8y+6yzā4z2
Then electric force acting on a 2C point charge placed at origin will be?
A
2 N
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B
6 N
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C
8 N
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D
20 N
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Solution
The correct option is D 20 N Ex=−dVdx=−(6−8y2),Ey=−dVdy=−(−16xy−8+6z),Ez=−dVdz=−(6y−8z)
At origin x=y=z=0 so, Ex=−6,Ey=8 and Ez=0 ⇒E=√E2x+E2y=10N/C
Hence force F=QE=2×10=20N