Question

Consider a fixed uniformly charged insulating sphere with radius \(R\) and total charge \(+Q\). A point charge \(-q\) (\(q \ll Q\)) with mass \(m\) is released from rest at a distance of \(3R\) from the centre of the charged sphere. When the point charge reaches the surface of the sphere, its speed is:

• A. \(\sqrt{\frac{Qq}{4\pi\epsilon_0 mR}}\)
• B. \(\sqrt{\frac{3Qq}{4\pi\epsilon_0 mR}}\)
• C. \(\sqrt{\frac{2Qq}{3\pi\epsilon_0 mR}}\)
• D. \(\sqrt{\frac{Qq}{3\pi\epsilon_0 mR}}\)

Hint:

Whenever a charged particle moves under electrostatic force only, use conservation of energy. Outside a uniformly charged sphere, \[ V=\frac{1}{4\pi\epsilon_0}\frac{Q}{r} \] just as for a point charge.

Answer 1

The Correct Option is C

Concept:

• Outside a uniformly charged sphere, the electric field and potential are the same as those of a point charge placed at its centre.

• Electrostatic force is conservative.

• Therefore mechanical energy remains conserved.

• Potential at distance \(r\) from the centre is \[ V=\frac{1}{4\pi\epsilon_0}\frac{Q}{r} \]

Step 1: Calculate the initial potential energy
Initially the charge is at \[ r_i=3R \] Potential at this point is \[ V_i=\frac{1}{4\pi\epsilon_0}\frac{Q}{3R} \] Since the moving charge is \(-q\), \[ U_i=(-q)V_i \] \[ U_i = -\frac{Qq}{12\pi\epsilon_0R} \]

Step 2: Calculate the final potential energy
At the surface, \[ r_f=R \] Potential at the surface is \[ V_f = \frac{1}{4\pi\epsilon_0} \frac{Q}{R} \] Hence \[ U_f = (-q)V_f \] \[ U_f = -\frac{Qq}{4\pi\epsilon_0R} \]

Step 3: Apply conservation of mechanical energy
Initially the particle is released from rest. Therefore, \[ K_i=0 \] Using \[ K_i+U_i = K_f+U_f \] we get \[ 0+U_i = \frac12 mv^2+U_f \] \[ \frac12 mv^2 = U_i-U_f \] \[ = -\frac{Qq}{12\pi\epsilon_0R} + \frac{Qq}{4\pi\epsilon_0R} \] \[ = \frac{Qq}{6\pi\epsilon_0R} \]

Step 4: Calculate the speed
\[ \frac12 mv^2 = \frac{Qq}{6\pi\epsilon_0R} \] \[ v^2 = \frac{Qq}{3\pi\epsilon_0mR} \] \[ v = \sqrt{\frac{2Qq}{3\pi\epsilon_0mR}} \] \[ \boxed{ v= \sqrt{\frac{2Qq}{3\pi\epsilon_0mR}} } \] Hence, \[ \boxed{\text{Option (C)}} \]