A parallel plate condenser of capacity 5`mu` F is kept connected to a battery of emf 10v. If the space between the plates is filled with a medium of dielectric constant 12, then the additional charge taken from the battery is

Options

A. `400 mu c`
B. `450 mu C `
C. `500 mu C `
D. `550 mu C `

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the initial charge on the capacitor
The initial charge (\(Q_i\)) on the capacitor can be calculated using the formula:
\[ Q_i = C \times V \]
where:
- \(C\) is the capacitance of the capacitor (5 microfarads, or \(5 \times 10^{-6} \, F\)),
- \(V\) is the voltage (10 volts).

Calculating \(Q_i\):
\[ Q_i = 5 \times 10^{-6} \, F \times 10 \, V = 50 \times 10^{-6} \, C = 50 \, \mu C \]

### Step 2: Calculate the new capacitance with the dielectric
When a dielectric material is introduced between the plates of the capacitor, the new capacitance (\(C'\)) can be calculated using the formula:
\[ C' = k \times C \]
where:
- \(k\) is the dielectric constant (12),
- \(C\) is the original capacitance (5 microfarads).

Calculating \(C'\):
\[ C' = 12 \times 5 \times 10^{-6} \, F = 60 \times 10^{-6} \, F = 60 \, \mu F \]

### Step 3: Calculate the final charge on the capacitor
The final charge (\(Q_f\)) on the capacitor after introducing the dielectric is given by:
\[ Q_f = C' \times V \]

Calculating \(Q_f\):
\[ Q_f = 60 \times 10^{-6} \, F \times 10 \, V = 600 \times 10^{-6} \, C = 600 \, \mu C \]

### Step 4: Determine the additional charge taken from the battery
The additional charge (\(Q_{additional}\)) taken from the battery when the dielectric is introduced can be calculated as:
\[ Q_{additional} = Q_f - Q_i \]

Calculating \(Q_{additional}\):
\[ Q_{additional} = 600 \, \mu C - 50 \, \mu C = 550 \, \mu C \]

### Final Answer
The additional charge taken from the battery is \(550 \, \mu C\).

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