A capacitor of capacitance `5muF` is charged to a potential difference 200 V and then allowed to dischage through a resistance `1kOmega`. The cahrge on the capacitor at the instant the current through the resistance is 100 Ma, is (in `muC`):

Options

A. 50
B. 100
C. 500
D. 1000

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Calculate the initial charge on the capacitor
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
where:
- \( C \) is the capacitance in farads,
- \( V \) is the voltage in volts.

Given:
- \( C = 5 \, \mu F = 5 \times 10^{-6} \, F \)
- \( V = 200 \, V \)

Substituting the values:
\[ Q = 5 \times 10^{-6} \, F \times 200 \, V = 1 \times 10^{-3} \, C = 1000 \, \mu C \]

### Step 2: Use the current to find the voltage across the resistor
The current \( I \) through the resistor is given as \( 100 \, mA = 100 \times 10^{-3} \, A \).

Using Ohm's Law:
\[ V_R = I \times R \]
where:
- \( R = 1 \, k\Omega = 1000 \, \Omega \)

Substituting the values:
\[ V_R = 100 \times 10^{-3} \, A \times 1000 \, \Omega = 100 \, V \]

### Step 3: Determine the voltage across the capacitor at the moment the current is 100 mA
Since the capacitor discharges through the resistor, the voltage across the capacitor \( V_C \) at that moment is equal to the voltage across the resistor:
\[ V_C = V_R = 100 \, V \]

### Step 4: Calculate the charge on the capacitor at that moment
Using the charge formula again:
\[ Q = C \times V_C \]
Substituting the values:
\[ Q = 5 \times 10^{-6} \, F \times 100 \, V = 5 \times 10^{-4} \, C = 500 \, \mu C \]

### Final Answer
The charge on the capacitor at the instant the current through the resistance is \( 100 \, mA \) is:
\[ \boxed{500 \, \mu C} \]