Three point charges $4q$, $Q$, and $q$ are placed in a straight line of length $L$ at points 0, $L/2$, and $L$ respectively. The net force on charge $q$ is zero. The value of $Q$ is:

Question

There are three co-centric conducting spherical shells $A$, $B$ and $C$ of radii $a$, $b$ and $c$ respectively $(c>b>a)$ and they are charged with charges $q_1$, $q_2$ and $q_3$ respectively. The potentials of the spheres $A$, $B$ and $C$ respectively are:

Answer 1

Step 1: Understanding the Concept
For the net force on the third charge ($q$) to be zero, the force exerted by the first charge ($4q$) and the second charge ($Q$) must be equal in magnitude but opposite in direction.
Step 2: Key Formula or Approach
Using Coulomb's Law: $ F = \frac{k q_1 q_2}{r^2} $. The net force on $q$ at position $L$ is: \[ F_{net} = F_{4q} + F_{Q} = 0 \]
Step 3: Detailed Calculation
1. Force due to $4q$ (at distance $L$): \[ F_{4q} = \frac{k (4q)(q)}{L^2} \] 2. Force due to $Q$ (at distance $L/2$): \[ F_{Q} = \frac{k (Q)(q)}{(L/2)^2} = \frac{4 k Q q}{L^2} \] 3. Set the sum to zero: \[ \frac{4kq^2}{L^2} + \frac{4kQq}{L^2} = 0 \] 4. Divide by common terms ($\frac{4kq}{L^2}$): \[ q + Q = 0 \implies Q = -q \]
Step 4: Final Answer
The value of Q is $ -q $.

Three point charges \(4q\), \(Q\), and \(q\) are placed in a straight line of length \(L\) at points 0, \(L/2\), and \(L\) respectively. The net force on charge \(q\) is zero. The value of \(Q\) is:

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The Correct Option is B

Solution and Explanation

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