Question

# The electric field components due to a charge inside the cube of side 0.1 m are shown in figure.where, $$E_x = \alpha$$, where $$\alpha$$ = 500 N/C-m, $$E_y$$ = 0, $$E_z = 0$$.Calculate(a) the flux through the cube and (b) the charge inside the cube.

Solution

## Given length of the cube=0.1 m$$E_y=0,\space E_z=0,\space E_x=500\dfrac{N}{C-m}$$the flux through the cube$$\psi=\int E.ds=\int (E_xi+E_yj+E_zk).ds=\int E_Xi.ds=E_xx=\alpha x=500x$$$$\psi={(E_x.A)}_{t=0.1}+{(E_x.A)}_{x=0.2}=E_xAcos180+E_xcos0=500\times0.01\times (-1)+500\times 0.2\times 0.01=-0.5+1=0.5\dfrac{N}{Cm^2}$$charge inside the cube$$\psi = \int E.ds=\dfrac{q}{E_0}\implies q=0.5\times E_0=0.5\times 8.85\times 10^{-12}=4.425\times 10^{-12}C$$Physics

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