Two identical metal spheres posses `+60C` and `-20C` of charges .They are brought in contact and then separated by `10cm` .the force between them is

Options

A. `36xx10^(13)N`
B. `36xx10^(14)N`
C. `36xx10^(12)N`
D. `3.6xx10^(12)N`

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Determine the initial charges on the spheres
Let the charges on the two spheres be:
- Sphere A: \( Q_1 = +60 \, \mu C \)
- Sphere B: \( Q_2 = -20 \, \mu C \)

### Step 2: Calculate the total charge when the spheres are brought into contact
When two identical metal spheres are brought into contact, they will share their total charge equally. The total charge \( Q_{total} \) is given by:
\[ Q_{total} = Q_1 + Q_2 = 60 \, \mu C + (-20 \, \mu C) = 40 \, \mu C \]

### Step 3: Calculate the charge on each sphere after contact
Since the spheres are identical, the charge on each sphere after they are brought into contact is:
\[ Q = \frac{Q_{total}}{2} = \frac{40 \, \mu C}{2} = 20 \, \mu C \]

### Step 4: Use Coulomb's Law to calculate the force between the spheres
Coulomb's Law states that the force \( F \) between two point charges is given by:
\[ F = k \frac{|Q_1' Q_2'|}{r^2} \]
where:
- \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, N \cdot m^2/C^2 \)
- \( Q_1' = 20 \, \mu C = 20 \times 10^{-6} \, C \)
- \( Q_2' = 20 \, \mu C = 20 \times 10^{-6} \, C \)
- \( r = 10 \, cm = 0.1 \, m \)

### Step 5: Substitute the values into the formula
Substituting the values into Coulomb's Law:
\[ F = 9 \times 10^9 \frac{(20 \times 10^{-6})(20 \times 10^{-6})}{(0.1)^2} \]
\[ F = 9 \times 10^9 \frac{400 \times 10^{-12}}{0.01} \]
\[ F = 9 \times 10^9 \times 40 \times 10^{-9} \]
\[ F = 360 \, N \]

### Final Answer
The force between the two spheres after being separated by 10 cm is \( 360 \, N \).

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