Problem 26 - A Complete Resource Book in Physics for JEE Main

Question

Three charges \(Q,+q\), and \(+q\) are placed at the vertices of a right angle triangle (isosceles triangle) as shown. The net electrostatic energy of the configuration is zero, if \(Q\) is equal to (A) \(\frac{-q}{1+\sqrt{2}}\) (B) \(\frac{-2 q}{2+\sqrt{2}}\) (C) \(-2 q\) (D) \(+q\)

Step-by-Step Solution

Answer

The charge Q has a value equal to: \[Q = \frac{-q}{1+\sqrt{2}}\] So, the correct answer is (A) \(\frac{-q}{1+\sqrt{2}}\).

Step 1: Identify the Potential Energy Interactions

In the given configuration, there are three charges: +q, +q, and Q. We'll have three potential energy interactions between these charges: - Interaction between charge Q and +q (let's call it U1) - Interaction between charge Q and another +q (let's call it U2) - Interaction between the two charges +q and +q (let's call it U3)

Step 2: Calculate the Potential Energy of the Interactions

The potential energy of the interaction between two point charges is given by the formula: \[ U = \frac{kq_1q_2}{r}\] where k is the Coulomb's constant, \(q_1\) and \(q_2\) are the interacting charges, and r is the distance between them. In our case, U1 and U2 will have equal potential energy interactions, as they involve the charges Q and +q with the same distance (lets call it d) between them. Therefore, \[U1 = U2 = \frac{kQq}{d}\] For charge interaction U3, there are two charges +q and +q, and the distance between them is the hypotenuse of the right-angled isosceles triangle. By Pythagorean theorem, the distance between these charges is \(d\sqrt{2}\). The potential energy of interaction U3 is then: \[U3 = \frac{kq^2}{d\sqrt{2}}\]

Step 3: Set the Net Electrostatic Energy to Zero

To set the net electrostatic energy of the configuration to zero, the sum of the three potential energy interactions U1, U2 and, U3 must be zero. Thus, \[U1 + U2 + U3 = 0\] Multiplying the energy equations with a factor 2, \[2U1 + 2U3 = \frac{-2kQq}{d} = \frac{-2kq^2}{d\sqrt{2}}\] Now, divide both sides by 2, \[\frac{kQq}{d} = \frac{-kq^2}{d\sqrt{2}}\] Canceling out k, q, and d from both sides, \[Q = \frac{-q}{1 + \sqrt{2}}\] Therefore, the charge Q has a value equal to: \[Q = \frac{-q}{1+\sqrt{2}}\] So, the correct answer is (A) \(\frac{-q}{1+\sqrt{2}}\).