A capacitor of capacitance `10 mu F` is charged to a potential `50 V` with a battery. The battery is now disconnected and an additional charge `200 mu C` is given to the positive plate of the capacitor. The potential difference across the capacitor will be.

Options

A. 50 V
B. 80 V
C. 100 V
D. 60 V

Explanation

To solve the problem step by step, we will follow the outlined process below:

### Step 1: Calculate the initial charge on the capacitor
The capacitance \( C \) of the capacitor is given as \( 10 \, \mu F \) (microfarads), and the potential difference \( V \) is \( 50 \, V \).

Using the formula for charge \( Q \):
\[ Q = C \times V \]
Substituting the values:
\[ Q = 10 \times 10^{-6} \, F \times 50 \, V = 500 \, \mu C \]

### Step 2: Add the additional charge to the positive plate
An additional charge of \( 200 \, \mu C \) is given to the positive plate of the capacitor. Therefore, the total charge on the positive plate becomes:
\[ Q_{total} = 500 \, \mu C + 200 \, \mu C = 700 \, \mu C \]

### Step 3: Determine the charge on the negative plate
Since the capacitor is isolated (the battery is disconnected), the charge on the negative plate remains unchanged at \( -500 \, \mu C \).

### Step 4: Calculate the net charge on the capacitor
The net charge on the capacitor is the charge on the positive plate minus the charge on the negative plate:
\[ Q_{net} = Q_{positive} - Q_{negative} = 700 \, \mu C - (-500 \, \mu C) = 700 \, \mu C + 500 \, \mu C = 1200 \, \mu C \]

### Step 5: Calculate the potential difference across the capacitor
Using the formula for potential difference \( V \):
\[ V = \frac{Q}{C} \]
Substituting the values:
\[ V = \frac{1200 \, \mu C}{10 \, \mu F} = \frac{1200 \times 10^{-6} \, C}{10 \times 10^{-6} \, F} = 120 \, V \]

### Final Answer
The potential difference across the capacitor will be \( 120 \, V \).

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