The objective is to determine the ratios \(\frac{V_2}{V_1}\) and \(\frac{U_2}{U_1}\) for two capacitors, with capacitances 2 \(\mu F\) and 3 \(\mu F\), connected in series to a V-volt battery.
The equivalent capacitance \(C_t\) for capacitors in series is calculated as:
\(\frac{1}{C_t} = \frac{1}{C_1} + \frac{1}{C_2}\)
Substituting the given capacitance values yields:
\(\frac{1}{C_t} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}\)
Therefore, the total capacitance is:
\(C_t = \frac{6}{5} \mu F\)
In a series connection, the charge \(Q\) is uniform across all capacitors. The voltage across each capacitor is determined by \(Q=CV\):
\(V_1 = \frac{Q}{C_1} \), \( V_2 = \frac{Q}{C_2}\)
Consequently, the voltage ratio is:
\(\frac{V_2}{V_1} = \frac{C_1}{C_2} = \frac{2}{3}\).
The energy stored in a capacitor is given by:
\(U = \frac{1}{2}CV^2\)
The ratio of energies stored is thus:
\(\frac{U_2}{U_1} = \frac{\frac{1}{2}C_2V_2^2}{\frac{1}{2}C_1V_1^2} = \frac{C_2V_2^2}{C_1V_1^2} = \frac{\frac{Q^2}{C_2}}{\frac{Q^2}{C_1}} = \frac{C_1}{C_2} = \frac{2}{3}\).
Both ratios, \(\frac{V_2}{V_1}\) and \(\frac{U_2}{U_1}\), are equal to \(\frac{2}{3}\), confirming the result:
\(\frac{V_2}{V_1} = \frac{U_2}{U_1} = \frac{2}{3}\)
A 10 $\mu\text{C}$ charge is placed in an electric field of $ 5 \times 10^3 \text{N/C} $. What is the force experienced by the charge?