Q.32. A parallel plate air capacitor has a capacitance C. When it is half filled as show in figure w

Q: What is the correct answer for this exam Physics Q32?

The correct answer is B. Sol. Initial capacitance $C = \frac{A \epsilon_0}{d} = C$. Final capacitance $C_{eq}$ when half filled with dielectric K=5. The capacitor can be considered as two capacitors in series. $C_1 = \frac{A \epsilon_0 K}{d/2} = \frac{A \epsilon_0 5}{d/2} = 10C$. $C_2 = \frac{A \epsilon_0}{d/2} = 2C$. For series combination, $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{10C} + \frac{1}{2C}

Q: Q.32. A parallel plate air capacitor has a capacitance C. When it is half filled as show in figure with a dielectric constant K = 5, the percentage increase in the capacitance is [Figure: A parallel…

The correct answer is B. Sol. Initial capacitance $C = \frac{A \epsilon_0}{d} = C$. Final capacitance $C_{eq}$ when half filled with dielectric K=5. The capacitor can be considered as two capacitors in series. $C_1 = \frac{A \epsilon_0 K}{d/2} = \frac{A \epsilon_0 5}{d/2} = 10C$. $C_2 = \frac{A \epsilon_0}{d/2} = 2C$. For series combination, $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{10C} + \frac{1}{2C}

Q: What are the options for this this exam Physics question?

A. 34 · B. 67 · C. 200 · D. 400

Question

Q.32. A parallel plate air capacitor has a capacitance C. When it is half filled as show in figure with a dielectric constant K = 5, the percentage increase in the capacitance is [Figure: A parallel plate capacitor with dielectric K=5 filling half the space between plates].

FirstInTest · Accepted Answer

Sol. Initial capacitance $C = \frac{A \epsilon_0}{d} = C$. Final capacitance $C_{eq}$ when half filled with dielectric K=5. The capacitor can be considered as two capacitors in series. $C_1 = \frac{A \epsilon_0 K}{d/2} = \frac{A \epsilon_0 5}{d/2} = 10C$. $C_2 = \frac{A \epsilon_0}{d/2} = 2C$. For series combination, $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{10C} + \frac{1}{2C} = \frac{1+5}{10C} = \frac{6}{10C} = \frac{3}{5C}$. So, $C_{eq} = \frac{5}{3}C$. Percentage increase $= \frac{C_{eq} - C}{C} 	imes 100 = \frac{\frac{5}{3}C - C}{C} 	imes 100 = \frac{\frac{2}{3}C}{C} 	imes 100 = 66.67$.

Answer

34

Answer

67

Answer

200

Answer

400

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