A steady potential of 500V is maintained across two capacitors, C1 = 5.0mf and C2 = 2.0mf connected in series. Calculate i) the potential difference across each capacitor, ii) the charge on each capacitor, iii) the energy stored in each capacitor.

For capacitors in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of individual capacitances.

$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$$ $$\frac{1}{C_{eq}} = \frac{1}{5.0 \times 10^{-3} \, \text{F}} + \frac{1}{2.0 \times 10^{-3} \, \text{F}}$$ $$\frac{1}{C_{eq}} = 200 \, \text{F}^{-1} + 500 \, \text{F}^{-1}$$ $$\frac{1}{C_{eq}} = 700 \, \text{F}^{-1}$$ $$C_{eq} = \frac{1}{700} \, \text{F}$$

In a series connection, the charge ($Q$) on each capacitor is the same and equal to the total charge stored by the equivalent capacitance.

$$Q = C_{eq} V$$ $$Q = \left(\frac{1}{700} \, \text{F}\right) \times (500 \, \text{V})$$ $$Q = \frac{500}{700} \, \text{C} = \frac{5}{7} \, \text{C}$$

The charge on each capacitor is $\boxed{\text{ii) } \frac{5}{7} \, \text{C} \approx 0.714 \, \text{C}}$.

The potential difference across each capacitor is given by $V = \frac{Q}{C}$.
For $C_1$:

$$V_1 = \frac{Q}{C_1}$$ $$V_1 = \frac{\frac{5}{7} \, \text{C}}{5.0 \times 10^{-3} \, \text{F}} = \frac{5}{7 \times 0.005} \, \text{V} = \frac{5}{0.035} \, \text{V} = \frac{1000}{7} \, \text{V}$$

For $C_2$:

$$V_2 = \frac{Q}{C_2}$$ $$V_2 = \frac{\frac{5}{7} \, \text{C}}{2.0 \times 10^{-3} \, \text{F}} = \frac{5}{7 \times 0.002} \, \text{V} = \frac{5}{0.014} \, \text{V} = \frac{2500}{7} \, \text{V}$$

The potential difference across each capacitor is:
$\boxed{\text{i) } V_1 = \frac{1000}{7} \, \text{V} \approx 142.86 \, \text{V}}$
$\boxed{\text{i) } V_2 = \frac{2500}{7} \, \text{V} \approx 357.14 \, \text{V}}$

The energy stored in a capacitor is given by $U = \frac{1}{2} Q V$.
For $C_1$:

$$U_1 = \frac{1}{2} Q V_1$$ $$U_1 = \frac{1}{2} \left(\frac{5}{7} \, \text{C}\right) \left(\frac{1000}{7} \, \text{V}\right)$$ $$U_1 = \frac{5000}{98} \, \text{J} = \frac{2500}{49} \, \text{J}$$

For $C_2$:

$$U_2 = \frac{1}{2} Q V_2$$ $$U_2 = \frac{1}{2} \left(\frac{5}{7} \, \text{C}\right) \left(\frac{2500}{7} \, \text{V}\right)$$ $$U_2 = \frac{12500}{98} \, \text{J} = \frac{6250}{49} \, \text{J}$$

The energy stored in each capacitor is:
$\boxed{\text{iii) } U_1 = \frac{2500}{49} \, \text{J} \approx 51.02 \, \text{J}}$
$\boxed{\text{iii) } U_2 = \frac{6250}{49} \, \text{J} \approx 127.55 \, \text{J}}$