JEE Advanced 2023 Paper 2 · Q01 · Electric Potential
An electric dipole is formed by two charges $+q$ and $-q$ located in the $x$-$y$ plane at $(0, 2)$ mm and $(0, -2)$ mm, respectively, as shown in the figure. The electric potential at point $P(100, 100)$ mm due to the dipole is $V_0$. The charges $+q$ and $-q$ are then moved to the points $(-1, 2)$ mm and $(1, -2)$ mm, respectively. What is the value of electric potential at $P$ due to the new dipole?
Reveal answer + step-by-step solution
Correct answer:B
Solution
The dipole potential at $\vec{r}$ is $V = \dfrac{k\,\vec{p}\cdot\hat{r}}{r^2}$ with $\vec{p}=q(\vec{r}_{+}-\vec{r}_{-})$. The observation point $P=(100,100)$ mm has $\hat{r}=\tfrac{1}{\sqrt 2}(\hat i+\hat j)$ and the same $r$ in both cases (the displacements are mm-scale, negligible at $P$), so the only change is in $\vec p\cdot\hat r$.
Original: $\vec p_1 = q[(0,2)-(0,-2)]\,\text{mm} = (0,\,4q)$ mm. $\vec p_1\cdot\hat r = \tfrac{4q}{\sqrt 2}$, so $V_0\propto \tfrac{4q}{\sqrt 2}$.
New: $\vec p_2 = q[(-1,2)-(1,-2)]\,\text{mm} = (-2q,\,4q)$ mm. $\vec p_2\cdot\hat r = \tfrac{-2q+4q}{\sqrt 2} = \tfrac{2q}{\sqrt 2}$.
Ratio: $\dfrac{V_{\text{new}}}{V_0} = \dfrac{2q/\sqrt 2}{4q/\sqrt 2} = \dfrac{1}{2}$, so $V_{\text{new}} = V_0/2$.
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