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Electrostatics question

2011 · Q97
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Electrostatics question

2011 · Q97

NEETPhysicsElectrostaticsMCQ+4 / −1
The electric potential V at any point (x, y, z), all in metres in space is given by V = 4x2 volt. The electric field at the point (1, 0, 2) in volt/meter, is
  1. A
    8 along negative X-axis
  2. B
    8 along positive X-axis
  3. C
    16 along negative X-axis
  4. D
    16 along positive X-axis
View written solutionFree

Correct answer: A

E→=−∇→V\overrightarrow E = - \overrightarrow \nabla VE=−∇V

where ∇→=i^∂∂x+j^∂∂y+∂∂z\overrightarrow \nabla = \widehat i{\partial \over {\partial x}} + \widehat j{\partial \over {\partial y}} + {\partial \over {\partial z}}∇=i∂x∂​+j​∂y∂​+∂z∂​

E→=−[i^∂V∂x+j^∂V∂y+k^∂V∂z]\overrightarrow E = - \left[ {\widehat i{{\partial V} \over {\partial x}} + \widehat j{{\partial V} \over {\partial y}} + \widehat k{{\partial V} \over {\partial z}}} \right]E=−[i∂x∂V​+j​∂y∂V​+k∂z∂V​]

Here, V = 4x2
∴E→=−8xi^ \therefore \overrightarrow E = - 8x\widehat i∴E=−8xi

The electric field at point (1, 0, 2) is

E→(1,0,2)=−8i^Vm−1{\overrightarrow E _{\left( {1,0,2} \right)}} = - 8\widehat iV{m^{ - 1}}E(1,0,2)​=−8iVm−1

So electric field is along the negative X-axis.

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