Two identical air capacitors with a capacity of C = 100 pF each are connected in series and connected

Two identical air capacitors with a capacity of C = 100 pF each are connected in series and connected to a current source with a voltage of U = 10.0 V how much will the charge change on each of the delta q capacitors if one of them is immersed in a dielectric with a dielectric constant epsilum = 2.0 ?

1. Find the initial battery charge.
Let’s write a formula for solving the problem:
q1 = C1 * U = C * U / 2;
Substitute the value into the formula:
q1 = 10 ^ -12 * 100/2;
q1 = 50 * 10 ^ -12 cells;
2. Find the new battery capacity.
Let’s write the formula for solving the problem:
1 / C2 = 1 / C + 1/2 * C = (2 + 1) / C;
We get:
C2 = C / 3;
Let’s combine the formulas:
q2 = C2 * U;
Let’s substitute:
q2 = 10 ^ -12 * 100/3;
q2 = 33.3 * 10 ^ -12 cells;
Δq = 16.67 * 10 ^ -12 cells.

Answer: Δq = 16.67 * 10 ^ -12 cells.