The potential due to an electrostatic charge | vedclass

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Question

(A) The electric field $E$ is given by the negative gradient of the potential: $E = -\frac{dV}{dr}$.Substituting $V(r) = \frac{q e^{-\alpha r}}{4 \pi

Vedclass · Accepted Answer

$2q / e$

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(A) The electric field $E$ is given by the negative gradient of the potential: $E = -\frac{dV}{dr}$.Substituting $V(r) = \frac{q e^{-\alpha r}}{4 \pi \varepsilon_{0} r}$,we get:$E = -\frac{d}{dr} \left( \frac{q e^{-\alpha r}}{4 \pi \varepsilon_{0} r} ight) = -\frac{q}{4 \pi \varepsilon_{0}} \left( \frac{r(-\alpha e^{-\alpha r}) - e^{-\alpha r}}{r^2} ight) = \frac{q e^{-\alpha r}}{4 \pi \varepsilon_{0} r^2} (1 + \alpha r)$.At $r = 1/\alpha$,the electric field is:$E(1/\alpha) = \frac{q e^{-1}}{4 \pi \varepsilon_{0} (1/\alpha)^2} (1 + \alpha(1/\alpha)) = \frac{q}{e} \cdot \frac{1}{4 \pi \varepsilon_{0} (1/\alpha^2)} \cdot 2 = \frac{2q \alpha^2}{4 \pi \varepsilon_{0} e}$.According to Gauss's Law,the flux $\phi$ through a sphere of radius $r$ is $\phi = E \cdot 4 \pi r^2 = \frac{q_{	ext{enclosed}}}{\varepsilon_{0}}$.Substituting $E$ at $r = 1/\alpha$:$\phi = \left( \frac{2q \alpha^2}{4 \pi \varepsilon_{0} e} ight) \cdot 4 \pi (1/\alpha)^2 = \frac{2q}{e \varepsilon_{0}}$.Equating this to $\frac{q_{	ext{enclosed}}}{\varepsilon_{0}}$,we find $q_{	ext{enclosed}} = 2q/e$.

The potential due to an electrostatic charge distribution is $V(r) = \frac{q e^{-\alpha r}}{4 \pi \varepsilon_{0} r}$,where $\alpha$ is positive. The net charge within a sphere centred at the origin and of radius $1/\alpha$ is

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