Figure given below 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 3. What will be the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric

The question doesn't provide a figure, but based on the description, we can still solve the problem.

Step 1: Understand the problem The problem is asking for the ratio of the total electrostatic energy stored in two identical parallel plate capacitors before and after the introduction of a dielectric.

Step 2: Recall the formula for electrostatic energy stored in a capacitor The electrostatic energy (U) stored in a capacitor is given by the formula U = 1/2 * C * V^2, where C is the capacitance and V is the voltage.

Step 3: Understand the effect of a dielectric on capacitance When a dielectric of dielectric constant K is introduced between the plates of a capacitor, the capacitance becomes K times the original capacitance. So, the new capacitance C' = K * C.

Step 4: Calculate the ratio of the total electrostatic energy before and after the introduction of the dielectric Before the dielectric is introduced, the total electrostatic energy is U1 = 1/2 * C * V^2. After the dielectric is introduced, the total electrostatic energy is U2 = 1/2 * K * C * V^2 = K * U1.

Therefore, the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric is U1/U2 = 1/K.

Given that the dielectric constant K = 3, the ratio U1/U2 = 1/3.