Question

The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

• A. \( \frac{\alpha}{r} \)
• B. \( \frac{\alpha r^2}{2} \)
• C. \( \frac{\alpha}{2r^2} \)
• D. \( -\frac{\alpha}{r} \)

Hint:

When the electric field is given in terms of \( r \), integrate it with respect to \( r \) to find the potential. The negative sign indicates that the potential decreases in the direction of the electric field.

Answer 1

The Correct Option is A

The relationship between electric field and potential is given by \( \vec{E} = -abla V \). The provided electric field is \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \). As the field is radial and dependent only on \( r \), we consider its radial component: \( E_r = \alpha \frac{1}{r^3} \). In one dimension, the electric field relates to the potential as \( E_r = -\frac{dV}{dr} \). Substituting the given electric field yields \( \alpha \frac{1}{r^3} = -\frac{dV}{dr} \). Integrating both sides with respect to \( r \) gives \( dV = -\alpha \frac{dr}{r^3} \). The integration results in \( V(r) = \int \alpha \frac{dr}{r^3} = \frac{\alpha}{2r^2} + C \), where \( C \) is the integration constant. Conventionally, the potential is set to zero at infinity, so \( C = 0 \). Therefore, the potential is \( V(r) = \frac{\alpha}{r} \). The magnitude of the potential is thus \( \frac{\alpha}{r} \), corresponding to option (A).