Question

The electric field at a point in a region is given by $\vec{E} = \frac{k}{r^2} \hat{r}$, where $k$ is a constant and $r$ is the distance of the point from the origin. The magnitude of potential of the point is

• A. $\frac{k}{r}$
• B. $kr$
• C. $-\frac{k}{r}$
• D. $-\frac{kr^2}{2}$

Answer 1

Answer: (A)

$\frac{k}{r}$

Answer 2

The relationship between the electric field $\vec{E}$ and electric potential $V$ is given by $\vec{E} = -\nabla V$. For a radially symmetric field, $\vec{E} = E_r \hat{r}$, this simplifies to $E_r = -\frac{dV}{dr}$. The given electric field is $\vec{E} = \frac{k}{r^2} \hat{r}$, so its radial component is $E_r = \frac{k}{r^2}$. Equating the two expressions for $E_r$: $\frac{k}{r^2} = -\frac{dV}{dr}$. Integrating both sides with respect to $r$ to find the potential $V$: $V = \int -\frac{k}{r^2} dr = -k \int r^{-2} dr = -k \left(\frac{r^{-1}}{-1}\right) + C = \frac{k}{r} + C$. Assuming the standard convention where potential is zero at infinity ($V(\infty) = 0$), we get $C=0$. Thus, the electric potential is $V = \frac{k}{r}$. The magnitude of the potential is $|V| = |\frac{k}{r}|$. Assuming $k$ is a positive constant (common in such problems), the magnitude is $\frac{k}{r}$.