Consider the system of capacitors shown in the figure below (C1 = 5.00 µF, C2 = 7.00 µF).

Find the equivalent capacitance of the system. Find the charge on each capacitor. Find the potential difference across each capacitor. Find the total energy stored by the group.

To find the equivalent capacitance of the system, we can use the following formula:

1/C_eq = 1/C1 + 1/C2

Plugging in the values, we get:

1/C_eq = 1/5.00 µF + 1/7.00 µF

Now, let's calculate the sum:

1/C_eq = 0.2 µF^(-1) + 0.1428 µF^(-1)

Adding these values gives:

1/C_eq = 0.3428 µF^(-1)

To get the equivalent capacitance, we simply take the reciprocal of both sides:

C_eq = 1 / (0.3428 µF^(-1))

C_eq = 2.916 µF

Therefore, the equivalent capacitance of the system is 2.916 µF.

To find the charge on each capacitor, we'll use the formula:

Q = C * V

First, calculate the charge on C1:

Q1 = C1 * V

Using the equivalent capacitance C_eq and the voltage across C1, we have:

Q1 = (5.00 µF) * V

Using the equivalent capacitance C_eq and the voltage across C2, we have:

Q2 = (7.00 µF) * V

Therefore, the charge on C1 is 5.00 µF times the voltage V, and the charge on C2 is 7.00 µF times the voltage V.

To find the potential difference (voltage) across each capacitor, we can use the formula:

V = Q / C

For C1:

V1 = Q1 / C1

Plugging in the values, we have:

V1 = (5.00 µF * V) / 5.00 µF

Simplifying, we find:

V1 = V

So, the potential difference across C1 is simply V.

Using the same formula, we can find the potential difference across C2:

V2 = Q2 / C2

Plugging in the values, we have:

V2 = (7.00 µF * V) / 7.00 µF

Simplifying, we find:

V2 = V

Similarly, the potential difference across C2 is also V.

Finally, to find the total energy stored by the group of capacitors, we can use the formula:

U = (1/2) * C * V^2

For the group of capacitors:

U_total = (1/2) * C_eq * V^2

Plugging in the value of C_eq, we have:

U_total = (1/2) * 2.916 µF * V^2

Simplifying, we find:

U_total = 1.458 µF * V^2

Therefore, the total energy stored by the group is 1.458 µF times the square of the voltage V.

Without "the figure below", we cannot help you with this. The capacitances add if the capacitors are in parallel. In series, the reciprocals add.