Question

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 as shown in the figure. The electric potential at point P(100, 100) mm due to dipole is V₀. Now the charges +q and -q are moved to (-1,2)mm and (1,-2)mm respectively. What is the value of the electric potential at point P due to this new dipole?

• A. \(\frac{V_0}{2}\)
• B. \(\frac{V_0}{4}\)
• C. \(\frac{V_0}{8}\)
• D. \(2V_0\)

Answer 1

The Correct Option is A

Electrostatic Potential Calculation

The correct option is: (A):

\(\vec{p_1} = q \times 4 \times 10^{-3} \hat{j}\)

Step 1: Calculate \(V_0\) at \((100,100) \, \text{mm}\)

The potential \(V_0\) is given by:

\(V_0 = \frac{K \vec{p_1} \cdot \vec{r_1}}{|\vec{r_1}|^3}\)

Substituting the known values:

\(V_0 = \frac{9 \times 10^9 \times q \times [4 \times 10^{-3} \hat{j}] \cdot (0.1 \hat{i} + 0.1 \hat{j})}{\left(\sqrt{(0.1)^2 + (0.1)^2}\right)^3}\)

Simplifying the dot product:

\(V_0 = \frac{9 \times 10^9 \times q \times [0.4 \times 10^{-3}]}{(0.1)^2 \times 2 \sqrt{2}}\)

Step 2: Moving Charges to New Positions

Now, +q and -q are moved to the points (-1,2) mm and (1,-2) mm, respectively. The new position vector is:

\(\vec{p} = q[-2 \hat{i} + 4 \hat{j}] \times 10^{-3}\)

The position vector \(\vec{r_1}\) remains:

\(\vec{r_1} = 0.1 \hat{i} + 0.1 \hat{j}\)

Step 3: Calculate the New Potential \(V\)

The new potential \(V\) is calculated as:

\(V = \frac{9 \times 10^9 (\vec{p}, \vec{r_1})}{|\vec{r_1}|^3}\)

Substituting the values:

\(V = \frac{9 \times 10^9 \times q \times [0.4 \times 10^{-3}]}{\left(\sqrt{(0.1)^2 + (0.1)^2}\right)^3}\)

Simplifying:

\(V = \frac{V_0}{2}\)

Thus, the new potential \(V\) is half of the original potential \(V_0\), confirming the relationship between the initial and final potentials.