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- 2k√6
- 2k√3
- k√6
- 4k√3

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Solution

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Given, V=k[2x2−y2+z2]

Electric field , →E=−(dVdx^i+dVdy^j+dVdz^k)

or,→E=−k(4x^i−2y^j+2z^k)

or,→E(1,1,1)=−k(4^i−2^j+2^k)

Magnitude of electric field=|→E(1,1,1)|=√k2(16+4+4)=k√24=2k√6

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