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
B
12mJ
C
12kJ
D
12μJ

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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