Problem 1: Capacitance of a Parallel-Plate Capacitor (No Dielectric)
A parallel-plate capacitor has plate area \(A = 0.02\,\text{m}^2\) and plate separation \(d = 1\,\text{mm} = 0.001\,\text{m}\). Calculate its capacitance in free space. Use \(\varepsilon_0 = 8.85 \times 10^{-12}\,\text{F/m}\).
Select the correct option:
Detailed Explanation
The capacitance of a parallel-plate capacitor is given by: \[ C = \frac{\varepsilon_0 A}{d}. \]
Substituting the values: \[ C = \frac{8.85 \times 10^{-12} \times 0.02}{0.001} = 1.77 \times 10^{-10}\,\text{F}. \]
Hence, the correct answer is Option A.
Problem 2: Capacitance with a Dielectric
A parallel-plate capacitor (with the same geometry as in Problem 1) is now filled with a dielectric of constant \(\kappa = 4\). Calculate its new capacitance.
Select the correct option:
Detailed Explanation
With a dielectric, the capacitance increases by a factor of \(\kappa\): \[ C' = \kappa C. \]
Hence, \[ C' = 4 \times 1.77 \times 10^{-10} = 7.08 \times 10^{-10}\,\text{F}. \]
Therefore, the correct answer is Option B.
Problem 3: Energy Stored in a Capacitor
A capacitor with capacitance \(C = 100\,\text{pF} = 100 \times 10^{-12}\,\text{F}\) is charged to a voltage \(V = 100\,\text{V}\). Calculate the energy stored in the capacitor.
Select the correct option:
Detailed Explanation
The energy stored in a capacitor is: \[ U = \frac{1}{2} C V^2. \]
Substituting: \[ U = \frac{1}{2} \times 100 \times 10^{-12} \times (100)^2 = \frac{1}{2} \times 100 \times 10^{-12} \times 10^4. \]
Simplify: \[ U = \frac{1}{2} \times 10^{-6} = 5 \times 10^{-7}\,\text{J}. \]
Thus, the correct answer is Option B.
Problem 4: Dielectric Insertion in an Isolated Capacitor
A capacitor of capacitance \(C_0\) is charged to a voltage \(V_0\) and then disconnected from the battery. If a dielectric with constant \(\kappa = 3\) is slowly inserted between the plates, what will be the new voltage across the capacitor?
Select the correct option:
Detailed Explanation
For an isolated (disconnected) capacitor the charge remains constant. Initially, \( Q = C_0 V_0 \). After insertion, the new capacitance is \( C' = \kappa C_0 \). The new voltage is: \[ V' = \frac{Q}{C'} = \frac{C_0 V_0}{\kappa C_0} = \frac{V_0}{\kappa}. \]
With \(\kappa = 3\), we obtain: \[ V' = \frac{V_0}{3}. \]
Hence, the answer is Option C.
Problem 5: Series Combination of Capacitors
Two capacitors of \(100\,\text{pF}\) and \(200\,\text{pF}\) are connected in series. Calculate their equivalent capacitance.
Select the correct option:
Detailed Explanation
For two capacitors in series, the equivalent capacitance \(C_{eq}\) is given by: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}. \]
Here, \(C_1 = 100\,\text{pF}\) and \(C_2 = 200\,\text{pF}\): \[ \frac{1}{C_{eq}} = \frac{1}{100} + \frac{1}{200} = 0.01 + 0.005 = 0.015\,\text{pF}^{-1}. \]
Therefore, \[ C_{eq} = \frac{1}{0.015} \approx 66.67\,\text{pF}. \]
So, the correct answer is Option B.
Problem 6: Parallel Combination of Capacitors
Two capacitors of \(150\,\text{pF}\) and \(250\,\text{pF}\) are connected in parallel. What is their equivalent capacitance?
Select the correct option:
Detailed Explanation
For capacitors in parallel, the equivalent capacitance is the sum: \[ C_{eq} = C_1 + C_2. \]
Hence, \[ C_{eq} = 150\,\text{pF} + 250\,\text{pF} = 400\,\text{pF}. \]
Thus, the correct answer is Option C.
Problem 7: Capacitance of a Different Parallel-Plate Capacitor
A parallel-plate capacitor has plate area \(A = 0.05\,\text{m}^2\) and plate separation \(d = 2\,\text{mm} = 0.002\,\text{m}\) in vacuum. Calculate its capacitance.
Select the correct option:
Detailed Explanation
The capacitance of a parallel-plate capacitor is: \[ C = \frac{\varepsilon_0 A}{d}. \]
Here, \(A = 0.05\,\text{m}^2\), \(d = 0.002\,\text{m}\), and \(\varepsilon_0 = 8.85 \times 10^{-12}\,\text{F/m}\). Thus: \[ C = \frac{8.85 \times 10^{-12} \times 0.05}{0.002} \approx 2.21 \times 10^{-10}\,\text{F}. \]
Hence, the answer is Option C.
Problem 8: Effect of Dielectric on Capacitance
When a dielectric with constant \(\kappa = 5\) is inserted completely between the plates of a capacitor, by what factor does the capacitance change?
Select the correct option:
Detailed Explanation
The presence of a dielectric increases the capacitance by a factor equal to the dielectric constant \(\kappa\). Therefore, the capacitance becomes: \[ C' = \kappa C. \]
With \(\kappa = 5\), the capacitance increases by a factor of 5.
So, the correct answer is Option C.
Problem 9: Energy Change upon Dielectric Insertion
A \(2\,\mu\text{F}\) capacitor is connected to a \(50\,\text{V}\) battery and then disconnected. If a dielectric of constant 2 is inserted, determine the new energy stored in the capacitor. (Hint: Use energy \(U = \frac{Q^2}{2C}\) with constant charge.)
Select the correct option:
Detailed Explanation
Initially, the capacitor’s charge is: \[ Q = CV = 2 \times 10^{-6}\,\text{F} \times 50\,\text{V} = 1 \times 10^{-4}\,\text{C}. \]
After insertion, the capacitance becomes \( C' = 2C = 4\,\mu\text{F}\). Since the capacitor is isolated, the charge remains constant.
The new energy stored is: \[ U' = \frac{Q^2}{2C'} = \frac{(1 \times 10^{-4})^2}{2 \times 4 \times 10^{-6}} = \frac{1 \times 10^{-8}}{8 \times 10^{-6}} = 1.25 \times 10^{-3}\,\text{J}. \]
Therefore, the correct answer is Option C.
Problem 10: Series Connection with Dielectric
Two identical capacitors, each of \(10\,\mu\text{F}\), are connected in series. A dielectric of constant 3 is then inserted across the series combination. What is the resulting equivalent capacitance?
Select the correct option:
Detailed Explanation
For two identical capacitors connected in series without a dielectric, the equivalent capacitance is: \[ C_{eq} = \frac{C}{2} = \frac{10\,\mu\text{F}}{2} = 5\,\mu\text{F}. \]
Inserting a dielectric with constant \(\kappa = 3\) increases the capacitance by that factor: \[ C'_{eq} = \kappa \times C_{eq} = 3 \times 5\,\mu\text{F} = 15\,\mu\text{F}. \]
Therefore, the correct answer is Option C.
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