MHT CET · Physics · Capacitance
Two identical capacitors A and B are connected in series to a battery of E.M.F., 'E'. Capacitor B contains a slab of dielectric constant \(\mathrm{K} . \mathrm{Q}_{\mathrm{A}}\) and \(\mathrm{Q}_{\mathrm{B}}\) are the charges stored in A and B . When the dielectric slab is removed, the corresponding charges are \(\mathrm{Q}_{\mathrm{A}}^{\prime}\) and \(\mathrm{Q}_{\mathrm{B}}\). Then
- A \(\frac{Q_A^{\prime}}{Q_A}=\frac{K}{2}\)
- B \(\frac{\mathrm{Q}_{\mathrm{B}}^{\prime}}{\mathrm{Q}_{\mathrm{B}}}=\frac{\mathrm{K}+1}{2}\)
- C \(\frac{Q_A^{\prime}}{Q_A}=\frac{K+1}{K}\)
- D \(\frac{\mathrm{Q}_{\mathrm{B}}^{\prime}}{\mathrm{Q}_{\mathrm{B}}}=\frac{\mathrm{K}+1}{2 \mathrm{~K}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{Q}_{\mathrm{B}}^{\prime}}{\mathrm{Q}_{\mathrm{B}}}=\frac{\mathrm{K}+1}{2 \mathrm{~K}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& C_A=C, C_B=K C \\
\therefore \quad & C_{n e t}=\frac{C_A C_B}{C_A+C_B}=\frac{C \times K C}{C+K C}=\left(\frac{K}{K+1}\right) C \\
& Q_A=Q_B=\frac{K C V}{K+1}
\end{aligned}\)
After the slab is removed,
\(C_A=C ; C_B=C\)
\(\begin{aligned}
\therefore \quad & \mathrm{C}_{\mathrm{net}}=\frac{\mathrm{C} \times \mathrm{C}}{\mathrm{C}+\mathrm{C}}=\frac{\mathrm{C}}{2} \\
& \mathrm{Q}_{\mathrm{A}}^{\prime}=\mathrm{Q}_{\mathrm{B}}^{\prime}=\frac{\mathrm{CV}}{2}
\end{aligned}\)
\(\therefore \quad \frac{\mathrm{Q}_{\mathrm{B}}^{\prime}}{\mathrm{Q}_{\mathrm{B}}}=\frac{\frac{\mathrm{CV}}{2}}{\frac{\mathrm{KCV}}{\mathrm{~K}+1}}=\frac{\mathrm{K}+1}{2 \mathrm{~K}}\)
& C_A=C, C_B=K C \\
\therefore \quad & C_{n e t}=\frac{C_A C_B}{C_A+C_B}=\frac{C \times K C}{C+K C}=\left(\frac{K}{K+1}\right) C \\
& Q_A=Q_B=\frac{K C V}{K+1}
\end{aligned}\)
After the slab is removed,
\(C_A=C ; C_B=C\)
\(\begin{aligned}
\therefore \quad & \mathrm{C}_{\mathrm{net}}=\frac{\mathrm{C} \times \mathrm{C}}{\mathrm{C}+\mathrm{C}}=\frac{\mathrm{C}}{2} \\
& \mathrm{Q}_{\mathrm{A}}^{\prime}=\mathrm{Q}_{\mathrm{B}}^{\prime}=\frac{\mathrm{CV}}{2}
\end{aligned}\)
\(\therefore \quad \frac{\mathrm{Q}_{\mathrm{B}}^{\prime}}{\mathrm{Q}_{\mathrm{B}}}=\frac{\frac{\mathrm{CV}}{2}}{\frac{\mathrm{KCV}}{\mathrm{~K}+1}}=\frac{\mathrm{K}+1}{2 \mathrm{~K}}\)
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