Two capacitors of capacitance `C` and `nC` are connected in parallel. A battery of emf `V` is connected across the combination. Now the battery is removed and a dielectric constant K is inserted filling the space between the plates of capacitor of capacitance C. The final potential difference across the system is
Two capacitors of capacitance `C` and `nC` are connected in parallel. A battery of emf `V` is connected across the combination. Now the battery is removed and a dielectric constant K is inserted filling the space between the plates of capacitor of capacitance C. The final potential difference across the system is
A
`(n+k)/(n+1)`
B
`(n+1)/(n+k)`
C
`(n-1)/(n+K)V`
D
`(n+1)/(n-k)V`
Text Solution
AI Generated Solution
The correct Answer is:
To solve the problem step by step, we will analyze the situation involving two capacitors connected in parallel and the effect of inserting a dielectric into one of them after disconnecting the battery.
### Step 1: Understand the Initial Setup
We have two capacitors:
- Capacitor 1: Capacitance \( C \)
- Capacitor 2: Capacitance \( nC \)
These capacitors are connected in parallel to a battery with an emf \( V \).
### Step 2: Calculate the Total Charge Before Removing the Battery
When the battery is connected, the total charge \( Q \) stored in the capacitors can be calculated using the formula:
\[
Q = C_{\text{eq}} \cdot V
\]
where \( C_{\text{eq}} \) is the equivalent capacitance of the capacitors in parallel. The equivalent capacitance for capacitors in parallel is given by:
\[
C_{\text{eq}} = C + nC = (1 + n)C
\]
Thus, the total charge \( Q \) is:
\[
Q = (1 + n)C \cdot V
\]
### Step 3: Remove the Battery
After the battery is removed, the total charge \( Q \) remains the same, but the capacitors are now isolated.
### Step 4: Insert the Dielectric
Now, a dielectric with a dielectric constant \( K \) is inserted into the capacitor with capacitance \( C \). The capacitance of this capacitor becomes:
\[
C' = K \cdot C
\]
The capacitance of the second capacitor remains \( nC \).
### Step 5: Calculate the New Equivalent Capacitance
The new equivalent capacitance \( C_{\text{eq}}' \) of the system after inserting the dielectric is:
\[
C_{\text{eq}}' = C' + nC = K \cdot C + nC = (K + n)C
\]
### Step 6: Calculate the Final Potential Difference
The final potential difference \( V' \) across the capacitors can be calculated using the charge \( Q \) and the new equivalent capacitance \( C_{\text{eq}}' \):
\[
V' = \frac{Q}{C_{\text{eq}}'}
\]
Substituting the values:
\[
V' = \frac{(1 + n)C \cdot V}{(K + n)C}
\]
The \( C \) cancels out:
\[
V' = \frac{(1 + n)V}{(K + n)}
\]
### Final Answer
Thus, the final potential difference across the system is:
\[
V' = V \cdot \frac{(1 + n)}{(K + n)}
\]
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