Two equally charged identical metal spheres X and Y repel each other with a force `5* 10^-4*` N another identical uncharged sphere C is touched to A and then placed at the mid-point between X and Y. Net electric force on C is

Options

A. `5x 10^(-4)N`
B. `15x 10^(-4)N`
C. `3x 10^(-4)N`
D. zero

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have two equally charged identical metal spheres, X and Y, that repel each other with a force of \( F = 5 \times 10^{-4} \, \text{N} \). An uncharged identical sphere, C, is touched to sphere X, acquiring half of its charge, and then placed at the midpoint between X and Y.

### Step 2: Determine the Charge on Spheres
Let the charge on each of the spheres X and Y be \( q \). The force of repulsion between them can be expressed using Coulomb's law:
\[ F = \frac{k \cdot q^2}{r^2} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \) is Coulomb's constant, \( q \) is the charge on each sphere, and \( r \) is the distance between the centers of the two spheres.

### Step 3: Charge Distribution on Sphere C
When sphere C is touched to sphere X, it acquires half of the charge from sphere X. Therefore, the charge on sphere C becomes:
\[ q_C = \frac{q}{2} \]
The charge on sphere X after touching sphere C becomes:
\[ q_X = \frac{q}{2} \]

### Step 4: Calculate the Forces on Sphere C
Now, sphere C is placed at the midpoint between spheres X and Y. The distance from sphere C to either sphere X or Y is \( \frac{r}{2} \).

#### Force on C due to X
The force \( F_{CX} \) on sphere C due to sphere X is given by:
\[ F_{CX} = \frac{k \cdot q_C \cdot q_X}{\left(\frac{r}{2}\right)^2} = \frac{k \cdot \left(\frac{q}{2}\right) \cdot \left(\frac{q}{2}\right)}{\left(\frac{r}{2}\right)^2} \]
\[ = \frac{k \cdot \frac{q^2}{4}}{\frac{r^2}{4}} = \frac{k \cdot q^2}{r^2} \]

#### Force on C due to Y
The force \( F_{CY} \) on sphere C due to sphere Y is given by:
\[ F_{CY} = \frac{k \cdot q_C \cdot q_Y}{\left(\frac{r}{2}\right)^2} = \frac{k \cdot \left(\frac{q}{2}\right) \cdot q}{\left(\frac{r}{2}\right)^2} \]
\[ = \frac{k \cdot \frac{q^2}{2}}{\frac{r^2}{4}} = \frac{2k \cdot q^2}{r^2} \]

### Step 5: Net Force on Sphere C
The net force \( F_C \) acting on sphere C is the vector sum of the forces due to spheres X and Y:
\[ F_C = F_{CY} - F_{CX} \]
Substituting the values we calculated:
\[ F_C = \frac{2k \cdot q^2}{r^2} - \frac{k \cdot q^2}{r^2} = \frac{k \cdot q^2}{r^2} \]

### Step 6: Relate to Given Force
From the earlier step, we know that:
\[ F = \frac{k \cdot q^2}{r^2} = 5 \times 10^{-4} \, \text{N} \]

### Conclusion
The net electric force on sphere C is:
\[ F_C = 5 \times 10^{-4} \, \text{N} \]