Question 245: Electrostatics & Capacitance

Question

For a short dipole placed at origin O , the dipole moment P is along $x$-axis, as shown in the figure. If the electric potential and electric field at $A$ are $V_0$ and $E_0$, respectively, then the correct combination of the electric potential and electric field, respectively, at point B on the $y$-axis is given by

Accepted Answer

Electric potential and field at point A:
At point A (on the $x$-axis), the electric field is $E_A = \frac{2kP}{r^3} = E_0$ and the potential is $V_A = \frac{kP}{r^2} = V_0$.
Electric potential and field at point B:
Point B is on the $y$-axis, which is at the same distance from the origin as point A, but in a direction perpendicular to the dipole moment.
The field at point B is given by $E_B = \frac{kP}{(2r)^3} = \frac{E_0}{16}$. This is because the distance from O to B is $2r$, and the formula for the field off the axis (on the equatorial line) is used.
The potential at point B is $V_B = \frac{k \vec{p} \cdot \hat{r}}{(2r)^2} = 0$, since $\vec{p}$ and $\hat{r}$ are perpendicular at B, so their dot product is zero.

Answer

$\frac{V_0}{2}$ and $\frac{E_0}{16}$

Answer

zero and $\frac{E_0}{8}$

Answer

$\mathrm{V}_0$ and $\frac{\mathrm{E}_0}{4}$

Answer

zero and $\frac{E_0}{16}$

Electrostatics & Capacitance

Choose the correct answer: