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 step by step, we will analyze the forces acting on the uncharged sphere C after it has been touched by sphere X and placed at the midpoint between spheres X and Y.

### Step 1: Understanding the Initial Setup
We have two identical charged spheres, X and Y, which repel each other with a force of \( F = 5 \times 10^{-4} \, \text{N} \). Let's denote the charge on each sphere as \( Q \) and the distance between them as \( R \).

### Step 2: Applying Coulomb's Law
According to Coulomb's law, the force between two charges is given by:

\[ F = \frac{1}{4\pi \epsilon_0} \frac{Q_1 Q_2}{R^2} \]

Since both spheres have the same charge \( Q \), we can write:

\[ F = \frac{1}{4\pi \epsilon_0} \frac{Q^2}{R^2} \]

Given that \( F = 5 \times 10^{-4} \, \text{N} \), we can use this information later to find the charge \( Q \).

### Step 3: Charging Sphere C
When the uncharged sphere C is touched by sphere X, it acquires half of the charge from X because both spheres are identical. Therefore, the charge on sphere C after touching X will be:

\[ Q_C = \frac{Q}{2} \]

### Step 4: Placing Sphere C at the Midpoint
Sphere C is then placed at the midpoint between spheres X and Y. The distances from C to both X and Y are:

\[ d_{XC} = d_{YC} = \frac{R}{2} \]

### Step 5: Calculating the Forces on Sphere C
Now, we will calculate the forces acting on sphere C due to spheres X and Y.

#### Force due to Sphere X on Sphere C (\( F_{XC} \)):
Using Coulomb's law:

\[ F_{XC} = \frac{1}{4\pi \epsilon_0} \frac{Q_C Q}{(d_{XC})^2} = \frac{1}{4\pi \epsilon_0} \frac{\left(\frac{Q}{2}\right) Q}{\left(\frac{R}{2}\right)^2} \]

This simplifies to:

\[ F_{XC} = \frac{1}{4\pi \epsilon_0} \frac{\frac{Q^2}{2}}{\frac{R^2}{4}} = \frac{2}{4\pi \epsilon_0} \frac{Q^2}{R^2} = 2F \]

Since \( F = 5 \times 10^{-4} \, \text{N} \):

\[ F_{XC} = 2 \times 5 \times 10^{-4} = 10 \times 10^{-4} \, \text{N} \]

#### Force due to Sphere Y on Sphere C (\( F_{YC} \)):
Similarly, the force between sphere C and sphere Y is:

\[ F_{YC} = \frac{1}{4\pi \epsilon_0} \frac{Q_C Q}{(d_{YC})^2} = \frac{1}{4\pi \epsilon_0} \frac{\left(\frac{Q}{2}\right) Q}{\left(\frac{R}{2}\right)^2} \]

This also simplifies to:

\[ F_{YC} = 2F = 10 \times 10^{-4} \, \text{N} \]

### Step 6: Determining the Net Force on Sphere C
The net force on sphere C will be the vector sum of the forces due to spheres X and Y. Since both forces are repulsive and act in opposite directions, we can write:

\[ F_{net} = F_{YC} - F_{XC} = 10 \times 10^{-4} - 5 \times 10^{-4} = 5 \times 10^{-4} \, \text{N} \]

### Final Answer
The net electric force on sphere C is:

\[ F_{net} = 5 \times 10^{-4} \, \text{N} \]