Question

If $V=x-y+xy+2zV$, find the electrostatic energy stored in a cube of side $2m$ centred at the origin.

• A. $0.236mJ$
• B. $0.236\mu J$
• C. $0.236nJ$
• D. $0.236pJ$

Answer 1

The correct option is C $0.236nJ$

$V=x-y+xy+2zV{E}_x=\frac{-d{V}_x}{dx},E_y=\frac{-d{V}_y}{dy},E_z=-\frac{d{V}_z}{dz}{E}_x=(-y-1)\hat{i}{E}_y=(1-x)\hat{j}{E}_z=-2\hat{k}E=\int {\int }_v\int 1/2{\epsilon }_0[{(1+y)}^2+{(1-x)}^2+4]=1/2{\epsilon }_0{\int }_{z=-1}^1{\int }_{y=-1}^1{\int }_{x=-1}^1[(1+y^2)+{(1-x)}^2+4]dxdy\frac{2{\epsilon }_0}{2}{\int }_{-1}^1{\int }_{y=-1}^1{y}^2+1+2y+6+x^2-2x={\epsilon }_0{\int }_{-1}^1{[\frac{y^3}{3}+y{x}^2+y^2-2xy+6y]}_{-1}^1={\epsilon }_0{\int }_{-1}^{1}2/3+2x^2-4x+12dx{\epsilon }_0(\frac{2}{3}\times 2+2+12.2){\epsilon }_0(26+\frac{4}{3})J={\epsilon }_0\frac{82}{3}J=241.9\times {10}^{-12}J=0.24nJ$