KCET · Physics · Electrostatics
In the situation shown in the diagram, magnitude, if \(q<<|Q|\) and \(r \gg a\). The net force on the free charge \(-q\) and net torque on it about \(O\) at the instant shown are respectively. ( \(p=2 a Q\) is the dipole moment)

- A \(\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^3} \hat{\mathbf{i}},-\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^2} \hat{\mathbf{k}}\)
- B \(\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^2} \hat{\mathbf{k}}, \frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^3} \hat{\mathbf{i}}\)
- C \(-\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^2} \hat{\mathbf{k}},-\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^3} \hat{\mathbf{i}}\)
- D \(\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^3} \hat{\mathbf{i}},+\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^2} \hat{\mathbf{k}}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^3} \hat{\mathbf{i}},+\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^2} \hat{\mathbf{k}}\)
Step-by-step Solution
Detailed explanation
Net force due to dipole on equatorial
line, \(F_{e q}=E \cdot q\)

\(=\frac{K \mathbf{p}}{r^3} \hat{\mathbf{i}} q=\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^3} \hat{\mathbf{i}}\)
Net torque; \(\tau=F \times r_{\perp}\)
\(=\frac{1}{4} \pi \frac{p q}{r^3} \hat{\mathbf{i}} \times r(\hat{\mathbf{j}})=\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^2} \hat{\mathbf{k}}\)
Hence, option (d) is correct.
line, \(F_{e q}=E \cdot q\)

\(=\frac{K \mathbf{p}}{r^3} \hat{\mathbf{i}} q=\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^3} \hat{\mathbf{i}}\)
Net torque; \(\tau=F \times r_{\perp}\)
\(=\frac{1}{4} \pi \frac{p q}{r^3} \hat{\mathbf{i}} \times r(\hat{\mathbf{j}})=\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^2} \hat{\mathbf{k}}\)
Hence, option (d) is correct.
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