A capacitor has a capacitance of `8.5muF`. How much charge must be removed so as to reduce the potential difference between its plates by 50V?

Answer

To solve the problem of how much charge must be removed from a capacitor with a capacitance of \(8.5 \, \mu F\) to reduce the potential difference between its plates by \(50 \, V\), we can follow these steps:

### Step 1: Understand the relationship between charge, capacitance, and voltage
The relationship between charge (\(Q\)), capacitance (\(C\)), and voltage (\(V\)) is given by the formula:
\[ Q = C \cdot V \]
Where:
- \(Q\) is the charge in coulombs,
- \(C\) is the capacitance in farads,
- \(V\) is the voltage in volts.

### Step 2: Define the initial conditions
Let the initial voltage across the capacitor be \(V_0\). The initial charge on the capacitor can be expressed as:
\[ Q_0 = C \cdot V_0 \]

### Step 3: Define the final conditions after removing charge
After removing charge \(Q\), the new voltage across the capacitor becomes:
\[ V = V_0 - 50 \]
The new charge on the capacitor can then be expressed as:
\[ Q = Q_0 - Q \]

### Step 4: Set up the equation for the new charge
Using the capacitance formula for the new voltage, we have:
\[ Q = C \cdot V = C \cdot (V_0 - 50) \]
Substituting \(Q_0\) into the equation gives:
\[ Q_0 - Q = C \cdot (V_0 - 50) \]

### Step 5: Substitute \(Q_0\) and simplify
Substituting \(Q_0 = C \cdot V_0\) into the equation:
\[ C \cdot V_0 - Q = C \cdot (V_0 - 50) \]
This simplifies to:
\[ Q = C \cdot 50 \]

### Step 6: Calculate the charge to be removed
Now we can substitute the value of \(C\):
\[ Q = 8.5 \, \mu F \cdot 50 \, V \]
Convert \(8.5 \, \mu F\) to farads:
\[ Q = 8.5 \times 10^{-6} \, F \cdot 50 \, V \]
Calculating this gives:
\[ Q = 425 \times 10^{-6} \, C = 425 \, \mu C \]

### Final Answer
The amount of charge that must be removed to reduce the potential difference by \(50 \, V\) is:
\[ \boxed{425 \, \mu C} \]

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