Potential Energy Change on Circular Path — Electrostatics Trick in 59 Sec 🔥

 

❓ Question

Two charges $q_1$ and $q_2$ are separated by a distance of 30 cm.
A third charge $q_3$, initially at C (as shown), is moved along a circular path of radius 40 cm from C to D.

If the change in potential energy of $q_3$ due to this movement is

$$\frac{q_3 K}{4\pi\varepsilon_0},$$

find the value of $K$.

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🖼️ Question Image

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✍️ Short Solution

This is a pure potential-energy difference problem.
👉 Path does NOT matter — only initial and final positions matter.
We compare the electric potential at C and D due to fixed charges $q_1, q_2$.

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🔹 Step 1 — Understand the Geometry (MOST IMPORTANT 💯)**

From the figure:

• $q_1$ at point A

• $q_2$ at point B

• $AB = 30\text{ cm}$
  

• $AC = 40\text{ cm}$ (vertical)

• Arc CD is a quarter circle of radius 40 cm, centred at A

So:

• $AD = 40\text{ cm}$
  

• $BD = AD - AB = 40 - 30 = 10\text{ cm}$
  

Distances:

Position	from $q_1$	from $q_2$
C	40 cm	$\sqrt{30^2 + 40^2} = 50$ cm
D	40 cm	10 cm

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🔹 Step 2 — Electric Potential at C

$$V_C = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_1}{40} + \frac{q_2}{50} \right)$$

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🔹 Step 3 — Electric Potential at D

$$V_D = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_1}{40} + \frac{q_2}{10} \right)$$

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🔹 Step 4 — Potential Difference (KEY STEP 🔥)**

$$\Delta V = V_D - V_C$$

The $q_1$ terms cancel:

$$\Delta V = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_2}{10} - \frac{q_2}{50} \right)$$
$$= \frac{1}{4\pi\varepsilon_0} \left( \frac{5q_2 - q_2}{50} \right) = \frac{4q_2}{50 \cdot 4\pi\varepsilon_0}$$
$$= \frac{1}{4\pi\varepsilon_0}\left(\frac{6q_2}{1}\right)$$

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🔹 Step 5 — Change in Potential Energy

$$\Delta U = q_3 \Delta V = \frac{q_3}{4\pi\varepsilon_0} \cdot 6q_2$$

Comparing with given form:

$$\frac{q_3 K}{4\pi\varepsilon_0}$$

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✅ Final Answer

$$\boxed{K = 6q_2}$$

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⭐ Golden JEE Insight

• Electrostatic work is path-independent

• Circular motion ≠ zero work (that’s only for magnetic field!)

• Always compute potential at start and end

🧠 One-line memory:

Potential energy change = charge × (final potential − initial potential)