RE-NEET 2026 Physics Electric Potential MCQ Question

To solve the problem, we need to analyze the motion of a point charge $-q$ that is released from a distance of $3R$ from the center of a uniformly charged insulating sphere of radius $R$ with total charge $+Q$.

Step 1: Understanding the Electric Field

For a uniformly charged insulating sphere, the electric field $E$ outside the sphere (at a distance $r$ greater than $R$) behaves as if all the charge were concentrated at the center. Thus, the electric field at a distance $r$ from the center is given by:

$$E = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{r^2}$$

Step 2: Calculating the Potential Energy

The point charge $-q$ experiences a force due to the electric field from the sphere. The potential energy $U$ of the charge $-q$ at a distance $r$ from the center of the sphere is given by:

$$U = -qV$$

where $V$ is the electric potential at distance $r$. The potential $V$ due to the sphere at a distance $r$ from its center is:

$$V = \frac{Q}{4\pi \epsilon_0 r}$$

Thus, the potential energy can be expressed as:

$$U = -q \cdot \frac{Q}{4\pi \epsilon_0 r}$$

Step 3: Energy Conservation

Since the charge is released from rest, we can apply the conservation of mechanical energy. The initial potential energy when the charge is at distance $3R$ is:

$$U_i = -q \cdot \frac{Q}{4\pi \epsilon_0 (3R)} = -\frac{qQ}{12\pi \epsilon_0 R}$$

As the charge moves towards the sphere and reaches its surface (distance $R$), the potential energy now becomes:

$$U_f = -q \cdot \frac{Q}{4\pi \epsilon_0 R} = -\frac{qQ}{4\pi \epsilon_0 R}$$

The kinetic energy $K$ of the charge at the moment it reaches the surface is given by:

$$K = \frac{1}{2} mv^2$$

Step 4: Setting Up the Energy Conservation Equation

Using conservation of energy:

$$K_i + U_i = K_f + U_f$$

Since the initial kinetic energy $K_i = 0$, we have:

$$0 + U_i = \frac{1}{2} mv^2 + U_f$$

Plugging in the potential energies:

$$-\frac{qQ}{12\pi \epsilon_0 R} = \frac{1}{2} mv^2 - \frac{qQ}{4\pi \epsilon_0 R}$$

Rearranging gives:

$$\frac{1}{2} mv^2 = -\frac{qQ}{12\pi \epsilon_0 R} + \frac{qQ}{4\pi \epsilon_0 R}$$

Step 5: Simplifying the Right Side

Combine the terms on the right:

$$\frac{1}{2} mv^2 = \left( \frac{qQ}{4\pi \epsilon_0 R} - \frac{qQ}{12\pi \epsilon_0 R} \right)$$

To combine these fractions, we find a common denominator (which is $12\pi \epsilon_0 R$):

$$\frac{1}{2} mv^2 = \frac{3qQ}{12\pi \epsilon_0 R} - \frac{qQ}{12\pi \epsilon_0 R} = \frac{2qQ}{12\pi \epsilon_0 R} = \frac{qQ}{6\pi \epsilon_0 R}$$

Step 6: Solve for $v^2$

Now, multiply both sides by 2:

mv^2 = \frac{2qQ}{6\pi \epsilon_0 R} = \frac{qQ}{3\pi \epsilon_0 R}$$ Dividing by $m$:

v^2 = \frac{qQ}{3m\pi \epsilon_0 R}$$

Finally, taking the square root gives us:

v = \sqrt{\frac{qQ}{3m\pi \epsilon_0 R}}$$ ### Conclusion The correct answer is