Question

Three identical square metal plates $M_1,M_2$ and $M_3$ of side 10 cm & 5 mm thick are arranged as shown in figure. The plates are separated by sheets of paper 0.5 mm thick & of dielectric constant 5. The outer plates are connected together & connected to lower potential while inner plate to higher potential terminal of a battery. Capacitance between the terminals of the battery is : $({ϵ}_0=8.9{10}^{-12}{C}^2/N-m^2)$

• A. $1780pF$
• B. $890pF$
• C. $445pF$
• D. $222.5pF$

Answer 1

Answer: (A)

$1780pF$

Here three plates will make two capacitors with capacitance $C=\frac{KA{ϵ}_0}{d}$

As the battery terminals for two capacitors are common so they are in parallel.

Thus, the equivalent capacitance across battery is:

$C_{eq}=C+C=2C=2\times \frac{KA{ϵ}_0}{d}=2\times \frac{5\left(0.1{)}^2\right(8.9\times {10}^{12})}{0.5\times {10}^{-3}}=1780\times {10}^{-12}F=1780pF$