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Question

An air capacitor of capacity c=10μf is connected to a constant voltage battery of 12V. Now the space between the plates is filled with a liquid of dielectric constant 5. The (additional) charge that flows now from the battery to the capacitor is:


A

480μC

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B

120μC

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C

240μC

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D

600μC

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Solution

The correct option is A

480μC


Step 1: Given

Capacity: C=10μf

Voltage: V=12V

Dielectric Constant: K=5

Step 2: Formula Used

q=CV where qis the charge, C is the capacitance and V is the potential difference.

In the presence of a dielectric liquid with dielectric constant K, the above formula becomes q=KCV.

Step 3: Find the additional charge

Calculate the initial charge by using the formula

qi=CV=10μf×12V=120μC

Since a dielectric liquid is added, the final charge will be the product of initial charge and dielectric constant. Calculate the final charge by multiplying the initial charge by the dielectric constant.

qf=Kqi=5×120μC=600μC

Calculate the additional charge by subtracting the initial charge from the final charge

q=qf-qi=600μC-120μC=480μC

Hence, the charge that flows now from the battery to the capacitor is 480μC. Hence, option A is correct.


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