Two similarly and equally charged identical metal spheres A and B repel each other with a force of `2xx10^(-5)N`. A third identical unchared sphere C is touched with A and then placed at the midpoint between A and B. Find the net electric force on C.

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the Setup
We have two identical metal spheres A and B, both positively charged and repelling each other with a force of \( F_{AB} = 2 \times 10^{-5} \, \text{N} \). A third identical uncharged sphere C is touched with sphere A and then placed at the midpoint between A and B.

### Step 2: Determine the Charge Distribution
When sphere C is touched with sphere A, the charge from A will redistribute between A and C. Since both spheres A and C are identical, the charge on each will be half of the original charge \( Q \) on A. Thus, after touching:
- Charge on A: \( Q_A = \frac{Q}{2} \)
- Charge on C: \( Q_C = \frac{Q}{2} \)

### Step 3: Define the Distances
Let the distance between spheres A and B be \( R \). Since C is placed at the midpoint, the distances from C to A and C to B are both:
- Distance \( AC = \frac{R}{2} \)
- Distance \( BC = \frac{R}{2} \)

### Step 4: Calculate the Forces on Sphere C
Using Coulomb's Law, we can calculate the forces acting on sphere C due to spheres A and B.

**Force between A and C (F_CA):**
\[ F_{CA} = k \frac{Q_A \cdot Q_C}{(AC)^2} = k \frac{\left(\frac{Q}{2}\right) \left(\frac{Q}{2}\right)}{\left(\frac{R}{2}\right)^2} = k \frac{\frac{Q^2}{4}}{\frac{R^2}{4}} = k \frac{Q^2}{R^2} \]

**Force between B and C (F_CB):**
Since sphere B retains the original charge \( Q \):
\[ F_{CB} = k \frac{Q_B \cdot Q_C}{(BC)^2} = k \frac{Q \cdot \left(\frac{Q}{2}\right)}{\left(\frac{R}{2}\right)^2} = k \frac{\frac{Q^2}{2}}{\frac{R^2}{4}} = 2k \frac{Q^2}{R^2} \]

### Step 5: Determine the Net Force on Sphere C
The net force \( F_{net} \) on sphere C is the difference between the forces due to spheres B and A:
\[ F_{net} = F_{CB} - F_{CA} = 2k \frac{Q^2}{R^2} - k \frac{Q^2}{R^2} = k \frac{Q^2}{R^2} \]

### Step 6: Relate the Net Force to the Given Force
From the problem, we know that:
\[ F_{AB} = k \frac{Q^2}{R^2} = 2 \times 10^{-5} \, \text{N} \]
Thus, the net force on sphere C is:
\[ F_{net} = 2 \times 10^{-5} \, \text{N} \]

### Step 7: Conclusion
The net electric force on sphere C is directed away from sphere B, as \( F_{CB} \) is greater than \( F_{CA} \).

### Final Answer
The net electric force on sphere C is \( 2 \times 10^{-5} \, \text{N} \) directed away from sphere B.

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