Question

Given that $E=\left(\right(3x^2+y)\hat{i}+x\hat{y})kV/m$, 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. $12\mu J$

Answer 1

The correct option is D $12\mu J$

$E=(({3x}^2+y)\hat{j}+x\hat{y})kV/m$

Work done $=-2\mu C$

$(0,5,0)$ to $(2,-1,0)$

$dw=\frac{qE}{dl}$

$=\frac{-2\times {10}^{-6}\times \{({3x}^2+y)\hat{i}+\hat{x}y\}}{\sqrt{{(2-0)}^2+{(-1-5)}^2+{(0+0)}^2}}=12\mu J$