Question

Figure given shows two identical parallel plate capacitors connected to a battery with switch S closed. The switch is now opened and the free space between the plate of capacitors is filled with a dielectric of dielectric constant $2$. What will be the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric?

• A. $2:1$
• B. $5:1$
• C. $4:5$
• D. $5:4$

Answer 1

Answer: (C)

$4:5$

When switch (S) is closed,
$U_1=\frac{1}{2}C{V}^2+\frac{1}{2}C{V}^2=C{V}^2$
Charge on B,Q=CV
When switch(S) is open and dielectric$(K=2)$ is inserted in both capacitors, potential of A remains same while charge on B remains same.
$U_1=\frac{1}{2}\cdot 2C\cdot V^2+\frac{Q^2}{2\cdot 2C}$
$=C{V}^2+\frac{C^2{V}^2}{4C}$
$=\frac{5}{4}C{V}^2$
$\frac{U_1}{U_2}=\frac{4}{5}$.