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Electrostatics question

2009 · Q143
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Electrostatics question

2009 · Q143

NEETPhysicsElectrostaticsMCQ+4 / −1
The electric potential at a point (x, y, z) is given by V = −-−x2y −-− xz3 + 4

The electric field at that point is
  1. A
    E→=i^2xy+j^(x2+y2)+k^(3xz−y2)\overrightarrow E = \widehat i2xy + \widehat j\left( {{x^2} + {y^2}} \right) + \widehat k\left( {3xz - {y^2}} \right)E=i2xy+j​(x2+y2)+k(3xz−y2)
  2. B
    E→=i^z3+j^xyz+k^z2\overrightarrow E = \widehat i{z^3} + \widehat jxyz + \widehat k{z^2}E=iz3+j​xyz+kz2
  3. C
    E→=i^(2xy−z3)+j^xy2+k^3z2x\overrightarrow E = \widehat i\left( {2xy - {z^3}} \right) + \widehat jx{y^2} + \widehat k3{z^2}xE=i(2xy−z3)+j​xy2+k3z2x
  4. D
    E→=i^(2xy+z3)+j^x2+k^3xz2\overrightarrow E = \widehat i\left( {2xy + {z^3}} \right) + \widehat j{x^2} + \widehat k3x{z^2}E=i(2xy+z3)+j​x2+k3xz2
View written solutionFree

Correct answer: D

The electric potential at a point,
V = –x2y – xz3 + 4.

The field
E→=−∇→V=−(∂V∂xi^+∂V∂yj^+∂V∂zk^)\overrightarrow E = - \overrightarrow \nabla V = - \left( {{{\partial V} \over {\partial x}}\widehat i + {{\partial V} \over {\partial y}}\widehat j + {{\partial V} \over {\partial z}}\widehat k} \right)E=−∇V=−(∂x∂V​i+∂y∂V​j​+∂z∂V​k)

∴E→=i^(2xy+z3)+j^x2+k^(3xz2) \therefore \overrightarrow E = \widehat i\left( {2xy + {z^3}} \right) + \widehat j{x^2} + \widehat k\left( {3x{z^2}} \right)∴E=i(2xy+z3)+j​x2+k(3xz2)

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