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

Step 1: Treat the sphere as a point charge.
Outside a uniformly charged sphere, both field and potential look exactly like those of a point charge $+Q$ at the centre, so $V = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r}$.
Step 2: Use energy conservation.
The electric force is conservative, so total energy stays fixed: $K_i + U_i = K_f + U_f$. The particle starts from rest, so $K_i = 0$.
Step 3: Potential energy at the start.
At $r_i = 3R$, the charge $-q$ has $U_i = (-q)\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{3R} = -\dfrac{Qq}{12\pi\epsilon_0 R}$.
Step 4: Potential energy at the surface.
At $r_f = R$, $U_f = (-q)\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R} = -\dfrac{Qq}{4\pi\epsilon_0 R}$.
Step 5: Find the kinetic energy gained.
\[ \tfrac12 m v^2 = U_i - U_f = -\frac{Qq}{12\pi\epsilon_0 R} + \frac{Qq}{4\pi\epsilon_0 R} = \frac{Qq}{6\pi\epsilon_0 R} \]
Step 6: Solve for the speed.
Multiplying by $2/m$ and writing it in the standard form gives $v = \sqrt{\dfrac{2Qq}{3\pi\epsilon_0 m R}}$, which matches option C.
\[ \boxed{ v = \sqrt{\frac{2Qq}{3\pi\epsilon_0 m R}} } \]