Question:
An electric dipole shown in the figure. Work done to move a charge particle of \(1\mu C\) from point Q to P is \(x \times 10^{-7} J\), then the value of \(x\) is:
An electric dipole shown in the figure. Work done to move a charge particle of \(1\mu C\) from point Q to P is \(x \times 10^{-7} J\), then the value of \(x\) is:
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In dipole problems, same \(r\) $\Rightarrow$ focus on \(\cos\theta\) difference.
Updated On: Apr 17, 2026
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Correct Answer: 1.8
Solution and Explanation
Concept:
Work done in moving charge:
\[
W = q (V_P - V_Q)
\]
Potential due to dipole:
\[
V = \frac{1}{4\pi \epsilon_0} \cdot \frac{p \cos\theta}{r^2}
\]
Step 1:Given data
•\(q = 1\mu C = 10^{-6}C\)
•\(r = 1\,cm = 10^{-2}m\)
•\(\theta_P = 60^\circ,\ \theta_Q = 120^\circ\)
Step 2:Potential difference \[ V_P - V_Q = \frac{k p}{r^2}(\cos60^\circ - \cos120^\circ) \] \[ = \frac{k p}{r^2}\left(\frac{1}{2} - (-\frac{1}{2})\right) = \frac{k p}{r^2} \]
Step 3:Substitute values From dipole: \(p = q \times 2l\) Using values from figure and \(k = 9 \times 10^9\): \[ W = 10^{-6} \cdot \frac{9 \times 10^9 \cdot p}{(10^{-2})^2} \] \[ W = 1.8 \times 10^{-7} \, J \] Conclusion \[ W = x \times 10^{-7} \Rightarrow x = \boxed{1.8} \]
Step 1:Given data
•\(q = 1\mu C = 10^{-6}C\)
•\(r = 1\,cm = 10^{-2}m\)
•\(\theta_P = 60^\circ,\ \theta_Q = 120^\circ\)
Step 2:Potential difference \[ V_P - V_Q = \frac{k p}{r^2}(\cos60^\circ - \cos120^\circ) \] \[ = \frac{k p}{r^2}\left(\frac{1}{2} - (-\frac{1}{2})\right) = \frac{k p}{r^2} \]
Step 3:Substitute values From dipole: \(p = q \times 2l\) Using values from figure and \(k = 9 \times 10^9\): \[ W = 10^{-6} \cdot \frac{9 \times 10^9 \cdot p}{(10^{-2})^2} \] \[ W = 1.8 \times 10^{-7} \, J \] Conclusion \[ W = x \times 10^{-7} \Rightarrow x = \boxed{1.8} \]
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