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(C) Point $A$ is on the axial line of the dipole at distance $r$. The electric potential $V_A$ and electric field $E_A$ are given by:$V_A = \frac{kP}{

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Point $A$ is on the axial line of the dipole at distance $r$. The electric potential $V_A$ and electric field $E_A$ are given by:$V_A = \frac{kP}{r^2} = V_0$$E_A = \frac{2kP}{r^3} = E_0$Point $B$ is on the equatorial line of the dipole at distance $2r$. The electric potential $V_B$ at any point on the equatorial line is zero because the potential is given by $V = \frac{k \vec{P} \cdot \hat{r}}{r^2}$,and for the equatorial line,the angle between $\vec{P}$ and the position vector $\vec{r}$ is $90^\circ$,so $\cos(90^\circ) = 0$.The electric field $E_B$ on the equatorial line at distance $d = 2r$ is given by:$E_B = \frac{kP}{d^3} = \frac{kP}{(2r)^3} = \frac{kP}{8r^3}$Since $E_0 = \frac{2kP}{r^3}$,we have $\frac{kP}{r^3} = \frac{E_0}{2}$.Substituting this into the expression for $E_B$:$E_B = \frac{1}{8} 	imes \left( \frac{kP}{r^3} ight) = \frac{1}{8} 	imes \frac{E_0}{2} = \frac{E_0}{16}$.Thus,the potential is zero and the electric field is $\frac{E_0}{16}$.

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