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. $480\mu C$
• B. $120\mu C$
• C. $240\mu C$
• D. $600\mu C$

Answer 1

The correct option is A

$480\mu C$

Step 1: Given

Capacity: $C=10\mu f$

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 charge

Calculate the initial charge by using the formula

$$\begin{gathered} q_i=CV \\ =10\mu f\times 12V \\ =120\mu C \end{gathered}$$

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.

$$\begin{gathered} q_f=K{q}_i \\ =5\times 120\mu C \\ =600\mu C \end{gathered}$$

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

$$\begin{gathered} q=q_f-q_i \\ =600\mu C-120\mu C \\ =480\mu C \end{gathered}$$

Hence, the charge that flows now from the battery to the capacitor is $480\mu C$. Hence, option A is correct.