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Electrostatics question

2014 · Q129
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Electrostatics question

2014 · Q129

NEETPhysicsElectrostaticsMCQ+4 / −1
In a region, the potential is represented by V(x, y, z) = 6x −-− 8xy −-− 8y + 6yz,   where VVV is in volts and x, y, z are in metres. The electric force experienced by a charge of 2 coulomb situated at point (1, 1, 1) is
  1. A
    656\sqrt 565​ N
  2. B
    30 N
  3. C
    24 N
  4. D
    4354\sqrt {35}435​ N
View written solutionFree

Correct answer: D

E→=−∂V∂xi^−∂V∂yj^−∂V∂zk^\overrightarrow E = - {{\partial V} \over {\partial x}}\widehat i - {{\partial V} \over {\partial y}}\widehat j - {{\partial V} \over {\partial z}}\widehat kE=−∂x∂V​i−∂y∂V​j​−∂z∂V​k

=−[(6−8y)i^+(−8x−8+6z)j^+(6y)k^] = - \left[ {\left( {6 - 8y} \right)\widehat i + \left( { - 8x - 8 + 6z} \right)\widehat j + \left( {6y} \right)\widehat k} \right]=−[(6−8y)i+(−8x−8+6z)j​+(6y)k]

At (1, 1, 1), E→=2i^+10j^−6k^\overrightarrow E = 2\widehat i + 10\widehat j - 6\widehat kE=2i+10j​−6k

⇒(E→)=22+102+62=140=235\Rightarrow \left( {\overrightarrow E } \right) = \sqrt {{2^2} + {{10}^2} + {6^2}} = \sqrt {140} = 2\sqrt {35}⇒(E)=22+102+62​=140​=235​

∴F=qE→=2×235=435\therefore F = q\overrightarrow E = 2 \times 2\sqrt {35} = 4\sqrt {35}∴F=qE=2×235​=435​

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