question_answer

Figure shows two identical capacitors \[{{C}_{1}}\] and \[{{C}_{2}},\] each of \[2\mu F\] capacitance, connected to a battery of 5 V. Initially switch S is closed.

After sometime, S is left open and dielectric slabs of dielectric constant K = 5 are used and inserted to fill completely the space between the plates of the two capacitors. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted?

Answer:

(i) When switch S is closed, then potential difference across each capacitor is 5V. Potential across the battery, \[{{V}_{1}}={{V}_{2}}=5V\] Capacitance of the capacitor, \[{{C}_{1}}={{C}_{2}}=2\mu F\] Charge on each capacitor, \[{{q}_{1}}={{q}_{2}}=CV\] \[=(2\mu F)\times (5V)=10\mu C\] When switch S is opened, then the potential difference across \[{{C}_{1}}\] remains \[5\,V,\] while the charge on capacitor \[{{C}_{2}}\] remains \[10\mu C.\] After insertion of dielectric between the plates of each capacitor, the new capacitance of each capacitor becomes \[C{{'}_{1}}=C{{'}_{2}}=5\times 2\mu F=10\mu F\] (ii) Now, charge on capacitor \[{{C}_{1}},\] \[q{{'}_{1}}C{{'}_{1}}{{V}_{1}}=5\times 2\times 5=50\mu C\] Charge on capacitor \[{{C}_{2}}\] remains the same, i.e. \[10\mu C.\] (iii) Potential difference across \[{{C}_{1}}\] remains the same, while potential difference across \[{{C}_{2}}\] is \[V_{2}^{'}=\frac{{{q}_{2}}}{C_{2}^{'}}=\frac{10}{5\times 2}=1\,V\]