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Electrostatics question

2015 · Q112
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Electrostatics question

2015 · Q112

NEETPhysicsElectrostaticsMCQ+4 / −1
If potential (in volts) in a region is expressed as V(x, y, z) = 6xy −-− y + 2yz, the electric field (in N/C) at point (1, 1, 0) is
  1. A
    −(2i^+3j^+k^)- \left( {2\widehat i + 3\widehat j + \widehat k} \right)−(2i+3j​+k)
  2. B
    −(6i^+9j^+k^)- \left( {6\widehat i + 9\widehat j + \widehat k} \right)−(6i+9j​+k)
  3. C
    −(3i^+5j^+3k^)- \left( {3\widehat i + 5\widehat j + 3\widehat k} \right)−(3i+5j​+3k)
  4. D
    −(6i^+5j^+2k^)- \left( {6\widehat i + 5\widehat j + 2\widehat k} \right)−(6i+5j​+2k)
View written solutionFree

Correct answer: D

Potential in a region V = 6xy – y + 2yz

As we know the relation between electric
potential and electric field is E→=−dVdx\overrightarrow E = {{ - dV} \over {dx}}E=dx−dV​

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

E→=[(6yi^+(6x−1+2z)j^+(2y)k^)]\overrightarrow E = \left[ {\left( {6y\widehat i + \left( {6x - 1 + 2z} \right)\widehat j + \left( {2y} \right)\widehat k} \right)} \right]E=[(6yi+(6x−1+2z)j​+(2y)k)]

E→(1,1,0)=−(6i^+5j^+2k^){\overrightarrow E _{\left( {1,1,0} \right)}} = - \left( {6\widehat i + 5\widehat j + 2\widehat k} \right)E(1,1,0)​=−(6i+5j​+2k)

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