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

We know that the electric field is the negative gradient of the potential: \[ \vec{E} = -\nabla V \] The given electric field is: \[ \vec{E} = \alpha \frac{\hat{r}}{r^3} \] Since the electric field is radial and only depends on \( r \), we consider the radial component of the electric field: \[ E_r = \alpha \frac{1}{r^3} \] The relationship between the electric field and the potential in one dimension is: \[ E_r = -\frac{dV}{dr} \] Substituting the given electric field: \[ \alpha \frac{1}{r^3} = -\frac{dV}{dr} \] Now, we integrate both sides with respect to \( r \): \[ dV = -\alpha \frac{dr}{r^3} \] Integrating both sides: \[ V(r) = \int \alpha \frac{dr}{r^3} = \frac{\alpha}{2r^2} + C \] Where \( C \) is the constant of integration. In the context of electrostatics, we usually set the potential to zero at infinity, implying \( C = 0 \). Therefore: \[ V(r) = \frac{\alpha}{r} \] Thus, the magnitude of the potential is \( \frac{\alpha}{r} \), which is option (A).