KCET · Physics · Capacitance
A parallel plate capacitor with air as the dielectric has capacitance \(C\). A slab of dielectric constant \(K\) and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be
- A \((K+3) \frac{C}{4}\)
- B \((K+2) \frac{C}{4}\)
- C \((K+1) \frac{C}{4}\)
- D \(\frac{K C}{4}\)
Answer & Solution
Correct Answer
(A) \((K+3) \frac{C}{4}\)
Step-by-step Solution
Detailed explanation
The condenser with air as the dielectric has capacitance
\(C_{1}=\frac{\varepsilon_{0}}{d}\left(\frac{3 A}{4}\right)=\frac{3 \varepsilon_{0} A}{3 d}\)
Similarly, the condenser with \(K\) as the dielectric constant has capacitance
\(C_{2}=\frac{\varepsilon_{0} K}{d}\left(\frac{A}{4}\right)=\frac{\varepsilon_{0} A K}{4 d}\)
Since, \(C_{1}\) and \(C_{2}\) are in parallel
\(\begin{aligned}C_{n e t} &=C_{1}+C_{2} \\&=\frac{3 \varepsilon_{0} A}{4 d}+\frac{\varepsilon_{0} A K}{4 d} \\&=\frac{\varepsilon_{0} A}{d}\left[\frac{3}{4}+\frac{K}{4}\right]=\frac{C}{4}(K+3)\end{aligned}\)
\(C_{1}=\frac{\varepsilon_{0}}{d}\left(\frac{3 A}{4}\right)=\frac{3 \varepsilon_{0} A}{3 d}\)
Similarly, the condenser with \(K\) as the dielectric constant has capacitance
\(C_{2}=\frac{\varepsilon_{0} K}{d}\left(\frac{A}{4}\right)=\frac{\varepsilon_{0} A K}{4 d}\)
Since, \(C_{1}\) and \(C_{2}\) are in parallel
\(\begin{aligned}C_{n e t} &=C_{1}+C_{2} \\&=\frac{3 \varepsilon_{0} A}{4 d}+\frac{\varepsilon_{0} A K}{4 d} \\&=\frac{\varepsilon_{0} A}{d}\left[\frac{3}{4}+\frac{K}{4}\right]=\frac{C}{4}(K+3)\end{aligned}\)
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