Question

# An air capacitor of capacity $c=10\mu f$ 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

$480\mu C$

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B

$120\mu C$

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C

$240\mu C$

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D

$600\mu C$

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Solution

## The correct option is A $480\mu C$ Step 1: GivenCapacity: $C=10\mu f$Voltage: $V=12V$Dielectric Constant: $K=5$Step 2: Formula Used$q=CV$ where $q$is 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 chargeCalculate the initial charge by using the formula${q}_{i}=CV\phantom{\rule{0ex}{0ex}}=10\mu f×12V\phantom{\rule{0ex}{0ex}}=120\mu C$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.${q}_{f}=K{q}_{i}\phantom{\rule{0ex}{0ex}}=5×120\mu C\phantom{\rule{0ex}{0ex}}=600\mu C$Calculate the additional charge by subtracting the initial charge from the final charge$q={q}_{f}-{q}_{i}\phantom{\rule{0ex}{0ex}}=600\mu C-120\mu C\phantom{\rule{0ex}{0ex}}=480\mu C$Hence, the charge that flows now from the battery to the capacitor is $480\mu C$. Hence, option A is correct.

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