Two tiny conducting spheres are identical any carry charges of `-20.0 mu C` and `+50.0 muC`. They are separated by a distance of 2.50 cm. The spheres are brought into contact and then separated to a distance of 2.50 cm. Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive or repulsive.

Options

A. `1.30 xx 10^(2) N`, attractive
B. `3.24 xx 10^(3) N`, attractive
C. `3.24 xx 10^(3) N`, repulsive
D. `1.44 xx 10^(4) N`, repulsive

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the initial charges on the spheres
The charges on the two spheres are:
- Sphere 1: \( q_1 = -20.0 \, \mu C = -20.0 \times 10^{-6} \, C \)
- Sphere 2: \( q_2 = +50.0 \, \mu C = +50.0 \times 10^{-6} \, C \)

### Step 2: Calculate the total charge when the spheres are brought into contact
When the two spheres are brought into contact, the total charge \( Q \) is the sum of the individual charges:
\[ Q = q_1 + q_2 = (-20.0 \times 10^{-6}) + (50.0 \times 10^{-6}) = 30.0 \times 10^{-6} \, C \]

### Step 3: Find the charge on each sphere after contact
Since the spheres are identical, the total charge will be equally distributed between them:
\[ q' = \frac{Q}{2} = \frac{30.0 \times 10^{-6}}{2} = 15.0 \times 10^{-6} \, C \]
Thus, after contact, both spheres will have a charge of \( +15.0 \, \mu C \).

### Step 4: Calculate the force between the spheres after separation
After being separated, the spheres are now both positively charged. The distance between them is \( r = 2.50 \, cm = 0.025 \, m \).

Using Coulomb's Law, the force \( F \) between two point charges is given by:
\[ F = k \frac{|q_1' \cdot q_2'|}{r^2} \]
where \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \).

Substituting the values:
\[ F = 9 \times 10^9 \frac{(15.0 \times 10^{-6})^2}{(0.025)^2} \]

Calculating the numerator:
\[ (15.0 \times 10^{-6})^2 = 225 \times 10^{-12} \, C^2 \]

Calculating the denominator:
\[ (0.025)^2 = 0.000625 \, m^2 \]

Now substituting these values into the force equation:
\[ F = 9 \times 10^9 \frac{225 \times 10^{-12}}{0.000625} \]
\[ F = 9 \times 10^9 \frac{225 \times 10^{-12}}{6.25 \times 10^{-4}} = 9 \times 10^9 \times 3.6 \times 10^{-7} \]
\[ F = 3.24 \times 10^3 \, N \]

### Step 5: Determine the nature of the force
Since both spheres have positive charges, the force between them is **repulsive**.

### Final Answer
The magnitude of the force that each sphere now experiences is \( 3.24 \times 10^3 \, N \) and the force is **repulsive**.

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