Two identical spheres carrying charges `-9 mu C` and `5 mu C` respectively are kept in contact and then separated from each other. Point out true statement from the following. In each sphere

Options

A. `1.25 xx 10^(13)` electrons are in deficit
B. `1.25 xx 10^(13)` electrons are in excess
C. `2.15 xx 10^(13)` electrons are in excess
D. `2.15 xx 10^(13)` electrons are in deficit

Explanation

To solve the problem of two identical spheres carrying charges of \(-9 \mu C\) and \(5 \mu C\) respectively, we will follow these steps:

### Step 1: Understand the Initial Charges
- Sphere A has a charge of \(-9 \mu C\).
- Sphere B has a charge of \(5 \mu C\).

### Step 2: Calculate the Total Charge
When the two spheres are brought into contact, the total charge is conserved. We can calculate the total initial charge:
\[ Q_{\text{total}} = Q_A + Q_B = -9 \mu C + 5 \mu C = -4 \mu C \]

### Step 3: Determine the Final Charge on Each Sphere
Since the spheres are identical and they will share the total charge equally when they are in contact, the final charge on each sphere can be calculated as follows:
\[ Q_{\text{final}} = \frac{Q_{\text{total}}}{2} = \frac{-4 \mu C}{2} = -2 \mu C \]

### Step 4: Analyze the Charge on Each Sphere
After separation, both spheres will have a charge of \(-2 \mu C\). This indicates that both spheres have an excess of electrons because they carry a negative charge.

### Step 5: Calculate the Number of Excess Electrons
To find the number of excess electrons on each sphere, we use the formula:
\[ Q = n \cdot e \]
where:
- \(Q\) is the charge on the sphere,
- \(n\) is the number of excess electrons,
- \(e\) is the charge of one electron (\(1.6 \times 10^{-19} C\)).

Rearranging the formula gives:
\[ n = \frac{Q}{e} = \frac{-2 \times 10^{-6} C}{-1.6 \times 10^{-19} C} = 1.25 \times 10^{13} \]

### Conclusion
Thus, each sphere has \(1.25 \times 10^{13}\) excess electrons.

### Final Answer
Both spheres have an excess of \(1.25 \times 10^{13}\) electrons.

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