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Question

Given that $$E = ((3x^2
+ y) \hat i + x \hat y) kV/m$$, find the work done in moving a $$- 2 \mu C$$ charge from $$(0, 5, 0)$$ to $$(2, -1 , 0)$$ by taking the path :
$$y=5-3x$$


A
12J
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B
12mJ
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C
12kJ
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D
12μJ
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Solution

The correct option is D $$12\mu J$$
$$E=\left( \left( { 3x }^{ 2 }+y \right) \hat { j } +x\hat { y }  \right) kV/m$$
Work done $$=-2\mu C$$
$$\left( 0,5,0 \right) $$ to $$\left( 2,-1,0 \right) $$
$$dw=\dfrac { qE }{ dl } $$
      $$=\dfrac { -2\times { 10 }^{ -6 }\times \left\{ \left( { 3x }^{ 2 }+y \right) \hat { i } +\hat { x } y \right\}  }{ \sqrt { { \left( 2-0 \right)  }^{ 2 }+{ \left( -1-5 \right)  }^{ 2 }+{ \left( 0+0 \right)  }^{ 2 } }  } =12\mu J$$

Physics

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