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GUJCET · Physics · Electrostatic Potential and Capacitance
Ten identical square metallic plates are arranged as shown in figure. Length of each plate is 1 . The equivalent capacitance of this arrangement would be ________.

- A \(\frac{3 \varepsilon_0 l^2}{2 d}\)
- B \(\frac{5 \varepsilon_0 l^2}{3 d}\)
- C \(\frac{3 \varepsilon_0 l^2}{d}\)
- D \(\frac{4 \varepsilon_0 l^2}{d}\)
Answer & Solution
Correct Answer
(B) \(\frac{5 \varepsilon_0 l^2}{3 d}\)
Step-by-step Solution
Detailed explanation
(B)
Plate 1,2 , plate 3,4 and plate 5 , 6 form a parallel plate capacitor.

Equivalent circuit
\(\begin{array}{l} C _1=\frac{\varepsilon_0 l^2}{3 d}, C _2=\frac{\varepsilon_0 l^2}{d}, C _3=\frac{\varepsilon_0 l^2}{3 d} \\ C = C _1+ C _2+ C _3 \\ C =\frac{\varepsilon_0 l^2}{3 d}+\frac{\varepsilon_0 l^2}{d}+\frac{\varepsilon_0 l^2}{3 d} \\ C =\frac{5 \varepsilon_0 l^2}{3 d}\end{array}\)
Plate 1,2 , plate 3,4 and plate 5 , 6 form a parallel plate capacitor.

Equivalent circuit
\(\begin{array}{l} C _1=\frac{\varepsilon_0 l^2}{3 d}, C _2=\frac{\varepsilon_0 l^2}{d}, C _3=\frac{\varepsilon_0 l^2}{3 d} \\ C = C _1+ C _2+ C _3 \\ C =\frac{\varepsilon_0 l^2}{3 d}+\frac{\varepsilon_0 l^2}{d}+\frac{\varepsilon_0 l^2}{3 d} \\ C =\frac{5 \varepsilon_0 l^2}{3 d}\end{array}\)
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