Question:
A solid sphere of radius \(R\) carries a positive charge \(Q\) distributed uniformly throughout its volume. A very thin hole is drilled through its centre. A particle of mass \(m\) and charge \(-q\) performs simple harmonic motion about the centre of the sphere in this hole. The frequency of oscillation is
A solid sphere of radius \(R\) carries a positive charge \(Q\) distributed uniformly throughout its volume. A very thin hole is drilled through its centre. A particle of mass \(m\) and charge \(-q\) performs simple harmonic motion about the centre of the sphere in this hole. The frequency of oscillation is
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Inside a uniformly charged sphere,
\[
E\propto x
\]
so the electrostatic restoring force is proportional to displacement, leading to simple harmonic motion.
Updated On: Jun 22, 2026
- \(\dfrac{1}{2\pi}\left[\dfrac{Qq}{4\pi\varepsilon_0 R^3m}\right]^{\frac12}\)
- \(\dfrac{1}{2\pi}\left[\dfrac{Qq}{4\pi\varepsilon_0 R^2m}\right]^{\frac12}\)
- \(\dfrac{1}{2\pi}\left[\dfrac{Q}{4\pi\varepsilon_0 mR^3}\right]^{\frac12}\)
- \(\dfrac{1}{2\pi}\left[\dfrac{Qq}{4\pi\varepsilon_0 mR}\right]^{\frac12}\)
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The Correct Option is A
Solution and Explanation
Step 1: Write the electric field inside a uniformly charged sphere.
For a uniformly charged solid sphere, electric field at a distance \(x\) from the centre is \[ E=\frac{1}{4\pi\varepsilon_0}\frac{Qx}{R^3} \] The field is directly proportional to displacement \(x\).
Step 2: Find the force on charge \(-q\).
Force on the particle is \[ F=-qE \] Substituting the value of \(E\), \[ F=-q\left(\frac{1}{4\pi\varepsilon_0}\frac{Qx}{R^3}\right) \] \[ F=-\frac{Qq}{4\pi\varepsilon_0R^3}x \] The negative sign shows that the force is directed towards the centre, hence it is a restoring force.
Step 3: Compare with the equation of S.H.M.
For simple harmonic motion, \[ F=-m\omega^2x \] Comparing with \[ F=-\frac{Qq}{4\pi\varepsilon_0R^3}x, \] we get \[ m\omega^2=\frac{Qq}{4\pi\varepsilon_0R^3} \] Thus, \[ \omega=\left(\frac{Qq}{4\pi\varepsilon_0R^3m}\right)^{\frac12} \]
Step 4: Find the frequency.
Frequency is related to angular frequency by \[ f=\frac{\omega}{2\pi} \] Therefore, \[ f= \frac{1}{2\pi} \left( \frac{Qq}{4\pi\varepsilon_0R^3m} \right)^{\frac12} \]
Step 5: Final conclusion.
Hence, the frequency of oscillation is \[ \boxed{ \frac{1}{2\pi} \left[ \frac{Qq}{4\pi\varepsilon_0R^3m} \right]^{\frac12} } \]
For a uniformly charged solid sphere, electric field at a distance \(x\) from the centre is \[ E=\frac{1}{4\pi\varepsilon_0}\frac{Qx}{R^3} \] The field is directly proportional to displacement \(x\).
Step 2: Find the force on charge \(-q\).
Force on the particle is \[ F=-qE \] Substituting the value of \(E\), \[ F=-q\left(\frac{1}{4\pi\varepsilon_0}\frac{Qx}{R^3}\right) \] \[ F=-\frac{Qq}{4\pi\varepsilon_0R^3}x \] The negative sign shows that the force is directed towards the centre, hence it is a restoring force.
Step 3: Compare with the equation of S.H.M.
For simple harmonic motion, \[ F=-m\omega^2x \] Comparing with \[ F=-\frac{Qq}{4\pi\varepsilon_0R^3}x, \] we get \[ m\omega^2=\frac{Qq}{4\pi\varepsilon_0R^3} \] Thus, \[ \omega=\left(\frac{Qq}{4\pi\varepsilon_0R^3m}\right)^{\frac12} \]
Step 4: Find the frequency.
Frequency is related to angular frequency by \[ f=\frac{\omega}{2\pi} \] Therefore, \[ f= \frac{1}{2\pi} \left( \frac{Qq}{4\pi\varepsilon_0R^3m} \right)^{\frac12} \]
Step 5: Final conclusion.
Hence, the frequency of oscillation is \[ \boxed{ \frac{1}{2\pi} \left[ \frac{Qq}{4\pi\varepsilon_0R^3m} \right]^{\frac12} } \]
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