Question

Four plates are arranged as shown in the diagram. If area of each plate is A and the distance between two neighbouring parallel plates is d, then the capacitance of this system between A and B will be

• A. $\frac { 4 \varepsilon _ { 0 } A } { d }$
• B. $\frac { 3 \varepsilon _ { 0 } A } { d }$
• C. $\frac { 2 \varepsilon _ { 0 } A } { d }$
• D. $\frac { \varepsilon _ { 0 } A } { d }$

Answer 1

Answer: (C)

$\frac { 2 \varepsilon _ { 0 } A } { d }$

Answer 2

To solve such type of problem following guidelines should be follows

Guideline 1. Mark the number (1,2,3……..) on the plates

Guideline 2. Rearrange the diagram as shown below

Guideline 3. Since middle capacitor having plates 2, 3 is short circuited so it should be eliminated from the circuit

Hence equivalent capacitance between A and B $C _ { A B } = 2 \frac { \varepsilon _ { 0 } A } { d }$