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If $\vec E = \frac{{{E_0}x}}{a}\hat i\,\left( {x - mt} \right)$ then flux through the shaded area of a cube is

$E_0a^2$
Zero
$E_0a^3$
$-E_0a^3$
Solution
$\overrightarrow{\mathrm{E}}=\frac{\mathrm{E}_{0}(\mathrm{a})}{\mathrm{a}} \hat{\mathrm{i}}=\mathrm{E}_{0} \hat{\mathrm{i}}$
Flux $=\mathrm{E}_{0}(\hat{\mathrm{i}}) \cdot \mathrm{a}^{2}(\hat{\mathrm{i}}) $
$=\mathrm{E}_{0} \mathrm{a}^{2} $
Similar Questions
In steady state heat conduction, the equations that determine the heat current $j ( r )$ [heat flowing per unit time per unit area] and temperature $T( r )$ in space are exactly the same as those governing the electric field $E ( r )$ and electrostatic potential $V( r )$ with the equivalence given in the table below.
Heat flow | Electrostatics |
$T( r )$ | $V( r )$ |
$j ( r )$ | $E ( r )$ |
We exploit this equivalence to predict the rate $Q$ of total heat flowing by conduction from the surfaces of spheres of varying radii, all maintained at the same temperature. If $\dot{Q} \propto R^{n}$, where $R$ is the radius, then the value of $n$ is