The `500 muF` capacitor is charged at a steady rate of `100 mu C//s`. The potential difference across the capacitor will be 10 V after an interval of

Options

A. 5 s
B. 25 s
C. 20 s
D. 50 s

Answer

To solve the problem, we need to find out how long it will take for a `500 µF` capacitor to reach a potential difference of `10 V` when it is charged at a steady rate of `100 µC/s`.

### Step-by-Step Solution:

1. **Understand the relationship between charge, capacitance, and potential difference:**
The formula relating charge (Q), capacitance (C), and potential difference (V) is given by:
\[ Q = C \times V \]
Where:
- \( Q \) is the charge in coulombs (C),
- \( C \) is the capacitance in farads (F),
- \( V \) is the potential difference in volts (V).

2. **Convert the capacitance to farads:**
Given:
\[ C = 500 \, \mu F = 500 \times 10^{-6} \, F = 5 \times 10^{-4} \, F \]

3. **Calculate the charge required for a potential difference of 10 V:**
Using the formula from step 1:
\[ Q = C \times V = (5 \times 10^{-4} \, F) \times (10 \, V) = 5 \times 10^{-3} \, C = 5000 \, \mu C \]

4. **Determine the time required to charge the capacitor:**
We know the charging rate is \( 100 \, \mu C/s \). We can use the formula:
\[ Q = \text{Rate} \times \text{Time} \]
Rearranging gives:
\[ \text{Time} = \frac{Q}{\text{Rate}} = \frac{5000 \, \mu C}{100 \, \mu C/s} \]

5. **Calculate the time:**
\[ \text{Time} = \frac{5000}{100} = 50 \, s \]

Thus, the potential difference across the capacitor will be `10 V` after an interval of **50 seconds**.

### Final Answer:
The time interval required is **50 seconds**.