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Question

# Electric potential in a region is varying according to the relation V=3x22−y24 , where x and y are in meter and V is in volt. Electric field intensity in N/C at a point (1 m, 2 m) is

A
3^i^j
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B
3^i+^j
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C
6^i2^j
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D
6^i+2^j
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Solution

## The correct option is B −3^i+^jElectric potential, V=3x22−y24 Relation between electric potential and electric field intensity is given by, →E=−[∂V∂x^i+∂V∂y^j+∂V∂z^k]....(1) Now, ∂V∂x=∂(3x22−y24)∂x=3x Similarly, ∂V∂y=−y2 and ∂V∂z=0 From (1) equation, we have →E=−3x^i+y2^j At point (1 m,2 m) →E=(−3×1)^i+22^j ∴→E=−3^i+^j Hence, option (b) is the correct answer.

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