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Question

Electric potential is given by V=6xāˆ’8xy2āˆ’8y+6yzāˆ’4z2
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=(68y2),Ey=dVdy=(16xy8+6z),Ez=dVdz=(6y8z)
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

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