Three identical charges are placed at corners of a equilateral triangel of side l. If force between any two charges is F, the work required to double the dimensiions of triangle is

Options

A. `-3Fl`
B. `3Fl`
C. `(-3/2) Fl`
D. `(3/2) Fl`

Answer

To solve the problem of finding the work required to double the dimensions of an equilateral triangle with identical charges at its corners, we will follow these steps:

### Step 1: Understand the Initial Setup
We have three identical charges \( Q \) placed at the corners of an equilateral triangle with side length \( l \). The force \( F \) between any two charges can be expressed using Coulomb's law:
\[ F = \frac{kQ^2}{l^2} \]
where \( k \) is Coulomb's constant.

### Step 2: Calculate the Initial Potential Energy
The potential energy \( U \) of the system of three charges can be calculated by considering the potential energy contributions from each pair of charges. There are three pairs of charges (1-2, 2-3, and 3-1):
\[ U = U_{12} + U_{23} + U_{31} = 3 \cdot \frac{kQ^2}{l} \]
Thus, the initial potential energy is:
\[ U_{\text{initial}} = 3 \cdot \frac{kQ^2}{l} \]

### Step 3: Determine the Final Setup After Doubling the Dimensions
When the dimensions of the triangle are doubled, the new side length becomes \( 2l \). The force between any two charges at this new distance is:
\[ F' = \frac{kQ^2}{(2l)^2} = \frac{kQ^2}{4l^2} \]
The potential energy of the system with the new side length is:
\[ U_{\text{final}} = 3 \cdot \frac{kQ^2}{2l} = \frac{3kQ^2}{2l} \]

### Step 4: Calculate the Work Done
The work done \( W \) by an external agent to change the configuration of the charges is equal to the change in potential energy:
\[ W = U_{\text{final}} - U_{\text{initial}} \]
Substituting the values we calculated:
\[ W = \frac{3kQ^2}{2l} - 3 \cdot \frac{kQ^2}{l} \]
This simplifies to:
\[ W = \frac{3kQ^2}{2l} - \frac{6kQ^2}{2l} = -\frac{3kQ^2}{2l} \]

### Step 5: Substitute \( kQ^2 \) in Terms of \( F \)
From the expression for force \( F = \frac{kQ^2}{l^2} \), we can express \( kQ^2 \) as:
\[ kQ^2 = Fl^2 \]
Substituting this into the work done expression:
\[ W = -\frac{3(Fl^2)}{2l} = -\frac{3Fl}{2} \]

### Final Answer
Thus, the work required to double the dimensions of the triangle is:
\[ W = -\frac{3Fl}{2} \]