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 $\mathrm{V}_0$ and $\mathrm{E}_0$, respectively, then the correct combination of the electric potential and electric field, respectively, at point $B$ on the $y$-axis is given by

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Aicharya · Accepted Answer

Given: - The electric potential of $A$ is $V_0$ - The electric field of $A$ is $E_0$. We need to determine the electric potential and electric field at point 8. Step El: Electric Potential of a Dipole The electric potential $V$ due to a short dipole at a point at distance $r$ is: $V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{p \cdot \hat{r}}{r^2}$ On the axial line (Point A at distance $r$): On the equatorial line (Point B at distance 2r): Since the dipole potential is given by: $V_A=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{p}{r^2}=V_0$ $V=\frac{1}{4 \pi c_0} \cdot \frac{p \cos \theta}{r^2}$ and on the equatorial line $\theta=90^{\circ} \Rightarrow \cos 90^{\circ}=0$, $V_g=0$ Thus, the electric potential at 8 is zero. Step 2: Electric Field of a Dipole The magnitude of the electric field at a distance $r$ from a dipole: On the axial line: $B_{\pi k u l}=\frac{1}{4 \pi c_0} \cdot \frac{2 p}{r^3}$ Given that of $A$ (on the axial line of $r$), the field is $E_0$ : $E_A=E_n=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 p}{r^1}$ On the equatorial line: $E_{\text {nquatiol }}=\frac{1}{4 \pi c_0} \cdot \frac{p}{r^3}$ At B (distance 2r): $E_s=\frac{1}{4 \pi c_0} \cdot \frac{p}{(2 r)^3}=\frac{1}{4 \pi c_0} \cdot \frac{p}{6 r^3}$ Since $E_0=\frac{1}{4 m_1} \cdot \frac{2 p}{p}$, we con express $E_n$ in terms of $E_0$ : $E_n=\frac{E_0}{16}$ Final Answer: - Electric potential at $\mathrm{E = 0}$ - Electric field ot $\mathrm{B}=\frac{N_y}{10}$

For a short dipole placed at origin $O$, the dipole moment $P$ is along $x$-axis, as shown in the figure. If…

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