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. 10 s
C. 50 s
D. 100 s

Answer

To solve the problem, we need to determine the time interval required for a 500 µF capacitor to reach a potential difference of 10 V when charged at a steady rate of 100 µC/s.

### Step-by-Step Solution:

1. **Understand the relationship between charge, capacitance, and potential difference:**
The potential difference (V) across a capacitor is given by the formula:
\[ V = \frac{Q}{C} \]
where \( Q \) is the charge on the capacitor and \( C \) is the capacitance.

2. **Identify the known values:**
- Capacitance, \( C = 500 \, \mu F = 500 \times 10^{-6} \, F \)
- Desired potential difference, \( V = 10 \, V \)
- Charging rate, \( \frac{dQ}{dt} = 100 \, \mu C/s = 100 \times 10^{-6} \, C/s \)

3. **Calculate the required charge (Q) for the desired potential difference:**
Rearranging the formula for potential difference, we can find the charge:
\[ Q = V \times C \]
Substituting the known values:
\[ Q = 10 \, V \times 500 \times 10^{-6} \, F = 5000 \, \mu C \]

4. **Determine the time required to accumulate this charge at the given rate:**
The charge accumulated over time \( t \) when charging at a rate of \( \frac{dQ}{dt} \) is given by:
\[ Q = \frac{dQ}{dt} \times t \]
Rearranging this gives:
\[ t = \frac{Q}{\frac{dQ}{dt}} \]
Substituting the values we have:
\[ t = \frac{5000 \, \mu C}{100 \, \mu C/s} = 50 \, s \]

5. **Conclusion:**
The time interval required for the capacitor to reach a potential difference of 10 V is **50 seconds**.

### Final Answer:
The potential difference across the capacitor will be 10 V after an interval of **50 seconds**.

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