Physics Q34 – NEET-UG 2026

Home Physics Electric Charges and FieldsQuestion #121

Consider a fixed uniformly charged insulating sphere with radius R and total charge +Q. A point charge –q (q<<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 : (

ϵ_{0}

is the permittivity of vacuum, neglect gravitational forces).

Clue 1 of 3: Conservation of Energy

The problem involves a charge moving in an electric field, and since only conservative electrostatic forces are acting (gravitational forces are neglected), the total mechanical energy (kinetic + potential) of the system is conserved.

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Step 1: Apply Conservation of Mechanical Energy✦ Active

Since gravitational forces are neglected and only the conservative electrostatic force acts, the total mechanical energy of the point charge is conserved. The initial kinetic energy is zero as the charge is released from rest.

K_{i} + U_{i} = K_{f} + U_{f}

Where $K_{i} = 0$ , $K_{f} = \frac{1}{2} m v^{2}$ , and $U = - q V (r)$ . The electric potential $V (r)$ outside a uniformly charged sphere is $V (r) = \frac{1}{4 π ϵ_{0}} \frac{Q}{r}$ .

Step 2: Calculate Initial and Final Potential Energies○ Expand

The initial position is $r_{i} = 3 R$ . The initial potential energy is:

U_{i} = - q V (3 R) = - q (\frac{1}{4 π ϵ_{0}} \frac{Q}{3 R}) = - \frac{Q q}{12 π ϵ_{0} R}

The final position is $r_{f} = R$ (surface of the sphere). The final potential energy is:

U_{f} = - q V (R) = - q (\frac{1}{4 π ϵ_{0}} \frac{Q}{R}) = - \frac{Q q}{4 π ϵ_{0} R}

💡 Teacher's Secret Hint

Remember that the potential energy of a negative charge in a positive potential is negative.

Step 3: Solve for the Final Speed○ Expand

Substitute the energy terms into the conservation of energy equation:

0 + U_{i} = \frac{1}{2} m v^{2} + U_{f}

Rearrange to solve for $v^{2}$ :

\frac{1}{2} m v^{2} = U_{i} - U_{f} = - \frac{Q q}{12 π ϵ_{0} R} - (- \frac{Q q}{4 π ϵ_{0} R})

\frac{1}{2} m v^{2} = \frac{Q q}{4 π ϵ_{0} R} - \frac{Q q}{12 π ϵ_{0} R}

\frac{1}{2} m v^{2} = \frac{3 Q q - Q q}{12 π ϵ_{0} R} = \frac{2 Q q}{12 π ϵ_{0} R} = \frac{Q q}{6 π ϵ_{0} R}

v^{2} = \frac{2 Q q}{6 π ϵ_{0} m R} = \frac{Q q}{3 π ϵ_{0} m R}

v = \sqrt{\frac{Q q}{3 π ϵ_{0} m R}}

This matches option (3).

💡 Teacher's Secret Hint

Ensure careful algebraic manipulation, especially with signs when subtracting potential energies.

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Physics Question 34 – NEET-UG 2026

Clue 1 of 3: Conservation of Energy

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