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PhyMedAdv 2023 • Paper 2

Electric Potential Due to Dipole Shifted in XY Plane

An electric dipole is formed by two charges +q+q and −q-q located in xyxy-plane at (0,2) mm(0,2)\text{ mm} and (0,−2) mm(0,-2)\text{ mm}, respectively, as shown in the figure. The electric potential at point P (100,100) mm\text{P }(100,100)\text{ mm} due to the dipole is V0V_0. The charges +q+q and −q-q are then moved to the points (−1,2) mm(-1,2)\text{ mm} and (1,−2) mm(1,-2)\text{ mm}, respectively. What is the value of electric potential at P\text{P} due to the new dipole?

Question Diagram 1

Options

A

V0/4V_0/4

B

V0/2V_0/2

Correct
C

V0/2V_0/\sqrt{2}

D

3V0/43V_0/4

Topics & Concepts

Electrostatic Potential and CapacitanceElectric Potential

Step-by-Step Solution

To find the electric potential at point P\text{P} due to the new dipole configuration, we can use the expression for the electric potential due to a short dipole at a distance r≫dr \gg d (where dd is the dipole length).

The electric potential VV at a position vector r⃗\vec{r} relative to the center of a dipole with dipole moment p⃗\vec{p} is given by: V=14πε0p⃗⋅r⃗r3V = \frac{1}{4\pi\varepsilon_0} \frac{\vec{p} \cdot \vec{r}}{r^3}

Step 1: Initial Dipole Configuration

The initial positions of the charges are:

  • Charge +q+q at (0,2) mm(0, 2)\text{ mm}
  • Charge −q-q at (0,−2) mm(0, -2)\text{ mm}

The center of this dipole is at the origin (0,0) mm(0, 0)\text{ mm}. The initial dipole moment vector p⃗1\vec{p}_1 is directed from −q-q to +q+q: p⃗1=q[(0−0)i^+(2−(−2))j^]=4qj^ mm⋅C\vec{p}_1 = q \left[ (0 - 0)\hat{i} + (2 - (-2))\hat{j} \right] = 4q\hat{j} \text{ mm}\cdot C

The position vector of point P\text{P} relative to the origin is: r⃗P=100i^+100j^ mm\vec{r}_P = 100\hat{i} + 100\hat{j} \text{ mm}

The dot product p⃗1⋅r⃗P\vec{p}_1 \cdot \vec{r}_P is: p⃗1⋅r⃗P=(4qj^)⋅(100i^+100j^)=400q mm2⋅C\vec{p}_1 \cdot \vec{r}_P = (4q\hat{j}) \cdot (100\hat{i} + 100\hat{j}) = 400q \text{ mm}^2\cdot C

Therefore, the initial potential V0V_0 at point P\text{P} is proportional to p⃗1⋅r⃗P\vec{p}_1 \cdot \vec{r}_P: V0=14πε0400qrP3V_0 = \frac{1}{4\pi\varepsilon_0} \frac{400q}{r_P^3}

Step 2: New Dipole Configuration

The new positions of the charges are:

  • Charge +q+q moved to (−1,2) mm(-1, 2)\text{ mm}
  • Charge −q-q moved to (1,−2) mm(1, -2)\text{ mm}

The center of the new dipole remains at the origin (−1+12,2−22)=(0,0) mm\left( \frac{-1+1}{2}, \frac{2-2}{2} \right) = (0, 0)\text{ mm}. The new dipole moment vector p⃗2\vec{p}_2 is: p⃗2=q[(−1−1)i^+(2−(−2))j^]=(−2i^+4j^)q mm⋅C\vec{p}_2 = q \left[ (-1 - 1)\hat{i} + (2 - (-2))\hat{j} \right] = (-2\hat{i} + 4\hat{j})q \text{ mm}\cdot C

The dot product p⃗2⋅r⃗P\vec{p}_2 \cdot \vec{r}_P is: p⃗2⋅r⃗P=(−2qi^+4qj^)⋅(100i^+100j^)=−200q+400q=200q mm2⋅C\vec{p}_2 \cdot \vec{r}_P = (-2q\hat{i} + 4q\hat{j}) \cdot (100\hat{i} + 100\hat{j}) = -200q + 400q = 200q \text{ mm}^2\cdot C

Step 3: Ratio of Potentials

The new electric potential V′V' at point P\text{P} is: V′=14πε0p⃗2⋅r⃗PrP3V' = \frac{1}{4\pi\varepsilon_0} \frac{\vec{p}_2 \cdot \vec{r}_P}{r_P^3}

Taking the ratio of V′V' to V0V_0: V′V0=p⃗2⋅r⃗Pp⃗1⋅r⃗P=200q400q=12\frac{V'}{V_0} = \frac{\vec{p}_2 \cdot \vec{r}_P}{\vec{p}_1 \cdot \vec{r}_P} = \frac{200q}{400q} = \frac{1}{2}

Thus, the potential at point P\text{P} due to the new dipole is: V′=V02V' = \frac{V_0}{2}

Correct Answer: (B) V0/2V_0/2

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