Question:
A parallel plate air capacitor has a capacitance \(C\). When it is half filled as shown in the figure with a dielectric constant \(K=5\), the percentage increase in the capacitance is:
A parallel plate air capacitor has a capacitance \(C\). When it is half filled as shown in the figure with a dielectric constant \(K=5\), the percentage increase in the capacitance is:
Updated On: Apr 10, 2026
- \(33.34\)
- \(66.67\)
- \(200\)
- \(400\)
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The Correct Option is B
Solution and Explanation
Concept:
When a dielectric slab fills part of the separation between plates (layered along thickness), the system behaves like capacitors in series.
Original capacitance:
\[
C=\frac{\varepsilon_0 A}{d}
\]
Here the dielectric fills half the thickness \(d/2\).
Step 1:Capacitance of the air region} \[ C_1=\frac{\varepsilon_0 A}{d/2} \] \[ C_1=\frac{2\varepsilon_0 A}{d} \]
Step 2:Capacitance of the dielectric region} \[ C_2=\frac{K\varepsilon_0 A}{d/2} \] \[ C_2=\frac{2K\varepsilon_0 A}{d} \]
Step 3:Equivalent capacitance} Since they are in series: \[ \frac{1}{C'}=\frac{1}{C_1}+\frac{1}{C_2} \] \[ \frac{1}{C'}= \frac{d}{2\varepsilon_0 A} + \frac{d}{2K\varepsilon_0 A} \] \[ \frac{1}{C'}=\frac{d}{2\varepsilon_0 A}\left(1+\frac{1}{K}\right) \] \[ C'=\frac{2K}{K+1}\frac{\varepsilon_0 A}{d} \] \[ C'=\frac{2K}{K+1}C \]
Step 4:Substitute \(K=5\)} \[ C'=\frac{10}{6}C \] \[ C'=\frac{5}{3}C \]
Step 5:Percentage increase} \[ \frac{C'-C}{C}\times100 \] \[ =\frac{\frac{5}{3}C-C}{C}\times100 \] \[ =\frac{2}{3}\times100 \] \[ =66.67% \] \[ \boxed{66.67} \]
Step 1:Capacitance of the air region} \[ C_1=\frac{\varepsilon_0 A}{d/2} \] \[ C_1=\frac{2\varepsilon_0 A}{d} \]
Step 2:Capacitance of the dielectric region} \[ C_2=\frac{K\varepsilon_0 A}{d/2} \] \[ C_2=\frac{2K\varepsilon_0 A}{d} \]
Step 3:Equivalent capacitance} Since they are in series: \[ \frac{1}{C'}=\frac{1}{C_1}+\frac{1}{C_2} \] \[ \frac{1}{C'}= \frac{d}{2\varepsilon_0 A} + \frac{d}{2K\varepsilon_0 A} \] \[ \frac{1}{C'}=\frac{d}{2\varepsilon_0 A}\left(1+\frac{1}{K}\right) \] \[ C'=\frac{2K}{K+1}\frac{\varepsilon_0 A}{d} \] \[ C'=\frac{2K}{K+1}C \]
Step 4:Substitute \(K=5\)} \[ C'=\frac{10}{6}C \] \[ C'=\frac{5}{3}C \]
Step 5:Percentage increase} \[ \frac{C'-C}{C}\times100 \] \[ =\frac{\frac{5}{3}C-C}{C}\times100 \] \[ =\frac{2}{3}\times100 \] \[ =66.67% \] \[ \boxed{66.67} \]
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