Two positive charges of magnitude q are placed at the ends of a side 1 of a square of side 2a. Two n

Q: Why is the final potential at the centre zero?

Because the centre is equidistant from all four corners. There are two positive charges and two negative charges of the same magnitude, so their potentials perfectly cancel each other out.

Q: How did we get the distance $\sqrt{5}a$?

The vertical distance from the top side to the bottom corners is $2a$, and the horizontal distance from the midpoint to the corner is $a$. Using the Pythagorean theorem, the distance is $\sqrt{(2a)^2 + a^2} = \sqrt{5a^2} = \sqrt{5}a$.

Q: Why do we use conservation of energy here?

The electrostatic force is a conservative force. Since no non-conservative forces (like friction) are doing work, the total mechanical energy (kinetic + potential) of the system remains constant.

Q: Does the mass of the charge $Q$ matter?

The mass of the charge $Q$ would matter if we needed to find its final velocity ($K = \frac{1}{2}mv^2$). However, since we only need the kinetic energy, the mass is not required.

Q: What if the charge $Q$ was negative?

If $Q$ was negative, its initial potential energy would be negative. It would gain potential energy as it moves to the centre (where potential is zero), meaning it would require external work to reach the centre from rest, or it wouldn't reach there naturally.

Question

Two positive charges of magnitude $q$ are placed at the ends of a side 1 of a square of side $2a$. Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge $Q$ moves from the middle of side 1 to the centre of square, its kinetic energy at the centre of square is

Options:

(A) $\frac{1}{4\pi\epsilon_0}\frac{2qQ}{a}\left(1-\frac{1}{\sqrt{5}}\right)$

(B) zero

(C) $\frac{1}{4\pi\epsilon_0}\frac{2qQ}{a}\left(1+\frac{1}{\sqrt{5}}\right)$

(D) $\frac{1}{4\pi\epsilon_0}\frac{2qQ}{a}\left(1-\frac{2}{\sqrt{5}}\right)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a square with a side length of $2a$. At the top two corners, we have positive charges $q$, and at the bottom two corners, we have negative charges $-q$.

Our protagonist, a charge $Q$, is initially placed exactly at the midpoint of the top side. It is released from rest and moves towards the centre of the square. We need to find its kinetic energy when it reaches the centre.

To solve this, we will use the principle of conservation of mechanical energy. Since the electrostatic force is conservative, the total mechanical energy of the system remains constant.

### The Initial Potential Energy

First, let's find the initial potential energy of the charge $Q$ at the midpoint of the top side. To do this, we need to calculate the electric potential $V_i$ at this point due to the four corner charges.

The distance from the midpoint to both the top $q$ charges is simply $a$.

The distance to the bottom $-q$ charges can be found using the Pythagorean theorem. The vertical distance is $2a$ and the horizontal distance is $a$, so the distance is $\sqrt{(2a)^2 + a^2} = \sqrt{5}a$.

Now, we sum the potentials:
$$V_i = \frac{kq}{a} + \frac{kq}{a} + \frac{k(-q)}{\sqrt{5}a} + \frac{k(-q)}{\sqrt{5}a}$$

Simplifying this, we get:
$$V_i = \frac{2kq}{a} - \frac{2kq}{\sqrt{5}a} = \frac{2kq}{a}\left(1 - \frac{1}{\sqrt{5}}\right)$$

The initial potential energy $U_i$ is simply the charge $Q$ multiplied by this potential:
$$U_i = Q V_i = \frac{2kqQ}{a}\left(1 - \frac{1}{\sqrt{5}}\right)$$

### The Journey to the Centre

Now, the charge $Q$ moves to the centre of the square. Let's calculate the final potential $V_f$ at the centre.

The distance from the centre to all four corners is equal, which is $\sqrt{a^2 + a^2} = \sqrt{2}a$.

We have two positive charges and two negative charges of the same magnitude at equal distances. Therefore, their potentials will perfectly cancel each other out:
$$V_f = \frac{kq}{\sqrt{2}a} + \frac{kq}{\sqrt{2}a} - \frac{kq}{\sqrt{2}a} - \frac{kq}{\sqrt{2}a} = 0$$

This means the final potential energy $U_f$ is also zero.

### Conservation of Energy

Finally, we apply the conservation of mechanical energy. The charge started from rest, so the initial kinetic energy $K_i$ is zero.

The loss in potential energy will exactly equal the gain in kinetic energy:
$$\Delta K + \Delta U = 0 \implies K_f - K_i = U_i - U_f$$

Substituting the values we found:
$$K_f - 0 = \frac{2kqQ}{a}\left(1 - \frac{1}{\sqrt{5}}\right) - 0$$

Replacing $k$ with $\frac{1}{4\pi\epsilon_0}$, we get our final answer:
$$K_f = \frac{1}{4\pi\epsilon_0}\frac{2qQ}{a}\left(1 - \frac{1}{\sqrt{5}}\right)$$

This is a beautiful demonstration of how energy conservation simplifies complex electrostatic problems!

Select Answer:

Visualized Solution

Visualizing the Setup

Initial Potential at Midpoint

Initial Potential Energy

Final Potential at Centre

Conservation of Energy

The Way Forward

The Sigma Insight: Electrostatic Potential Energy

Analyzing the Setup

The Initial Potential Energy

The Journey to the Centre

Conservation of Energy

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Select Answer:

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The Sigma Insight: Electrostatic Potential Energy

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