As with resistors, capacitors in series share the same current. This is a re-statement of the law of conservation of charge. In contrast, capacitors and resistors in parallel share the same voltage.

We are asked how much charge is stored in total on the circuit. We can use the equivalent capacitance and the voltage supplied by the battery to calculate the charge. Remember that Q = CV, where Q is the total charge, C is the equivalent capacitance, and V is the voltage. We must first solve for equivalent capacitance.

C₂ and C₃ are capacitors in series, while C₁ is in parallel.

\(\displaystyle \frac{1}{C_{23}}=\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{1\mu F}+\frac{1}{1\mu F}=2\mu F\)

C₂₃ = 0.5μF

C_eq = C₂₃ + C₁ = 0.5μF + 0.5μF = 1μF

Now we can plug in the C_eq and battery voltage to find the charge.

Q = (1μF)(10V) = 10μC

This question asks us to consider electromagnetism and what happens when a capacitor is charging. When a capacitor is charging, charge is accumulating on the surface over a period of time. Given that the electric field due to a capacitor is given by the formula E = σ/ε₀, where σ is the charge per unit area and ε₀ is the constant of permeability of free space, we can see that E is directly proportional to σ; therefore, the more charge that builds up, the higher the resulting E field.

Because σ is changing during charging, and thus E is changing, we also know that a magnetic field, B, must be created.

Remember the principle—a changing electric field induces a changing magnetic field, and a changing magnetic field induces a changing electric field.

First, we need to determine the type of energy being stored by the capacitors in the circuit. As this is an electric circuit, electric energy is being stored. Thus, we can use the formula for potential energy stored in a capacitor, U = ½CV². We must first find the equivalent capacitance.

C₂ and C₃ are capacitors in series, while C₁ is in parallel.

\(\displaystyle \frac{1}{C_{23}}=\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{1\mu F}+\frac{1}{1\mu F}=2\mu F\)

C₂₃ = 0.5μF

C_eq = C₂₃ + C₁ = 0.5μF + 0.5μF = 1μF

U = ½CV² = ½(1μF)(10V)² = 5 * 10^-5J 

Relevant equations:

\(\displaystyle Q = CV\)

\(\displaystyle C = \frac{\kappa \epsilon _o A}{d}\)

The first equation shows that charge \(\displaystyle Q\) is directly proportional to capacitance \(\displaystyle C\), so we want to maximize \(\displaystyle C\) in order to find the maximum charge. 

The second equation shows that \(\displaystyle C\) is directly proportional to the dielectric constant \(\displaystyle k\), and inversely proportional to distance \(\displaystyle d\). To increase capacitance, we need \(\displaystyle k\) to be large and \(\displaystyle d\) to be small.

A smaller distance \(\displaystyle \left ( \frac{d}{2}\right )\) and a larger dielectric constant (choosing glass instead of air) will lead to a large stored charge.

To solve this question, we need to use the equation that describes voltage decay during capacitor discharging:

\(\displaystyle V=V_o e^{\frac{-t}{RC}}\)

Here, \(\displaystyle V\) is the voltage after a certain amount of time, \(\displaystyle V_o\) is the initial voltage, \(\displaystyle t\) is the time elapsed, \(\displaystyle R\) is the resistance of the resistor, and \(\displaystyle C\) is the capacitance of the capacitor. We can rearrange this equation in such a way that we solve for capacitance, \(\displaystyle C\):

\(\displaystyle V=V_o e^(\frac{-t}{RC})\)

\(\displaystyle \frac{V}{V_o}=e^(\frac{-t}{RC})\)

To remove the exponential \(\displaystyle e\) from the equation, we need to take the natural logarithm of both sides:

\(\displaystyle ln (\frac{V}{V_o})= \frac{-t}{RC}\)

Solving for \(\displaystyle C\) gives us:

\(\displaystyle C= \frac{-t}{R\times ln\frac{V}{V_o}}\)

\(\displaystyle C= \frac{-2s}{3\Omega \times ln\frac{6V}{8V}}\)

\(\displaystyle C=2.32F\)

The time constant, \(\displaystyle \tau\), of an RC circuit is the product of the resistance and the capacitance; therefore, the time constant of this circuit is:

\(\displaystyle \tau =RC=3\Omega \times 2.32F=6.96s\)

A capacitor is an electrical device consisting of two parallel conducting plates that store charge. A capacitor undergoes two processes: charging and discharging. During charging, a capacitor accumulates and stores charge between the two plates. During discharge, a capacitor loses the stored charge. Since there is a decrease in the amount of charge during discharge, there is also a decrease (or decay) in current and voltage in a discharging capacitor. The decay of all these parameters is characterized by the equation:

\(\displaystyle A=A_oe^{\frac{-t}{RC}}\)

Here, \(\displaystyle A\) and \(\displaystyle A_o\) denote the variable (voltage, current, or charge), \(\displaystyle t\) denotes the time elapsed, \(\displaystyle R\) denotes the resistance of the resistor, and \(\displaystyle C\) denotes the capacitance of the capacitor. This is an equation for exponential decay; therefore, all the variables undergo exponential decay in a discharging capacitor. 

The question states that the potential difference dropped from \(\displaystyle 12V\) to \(\displaystyle 4.42V\) in \(\displaystyle 4ms\). The percentage of voltage drop is:

\(\displaystyle \frac{12V-4.42V}{12V}\times100 =63.2 \%\)

This means that there was a \(\displaystyle 63.2\%\) voltage drop in \(\displaystyle 4ms\). The definition of time constant is the time it takes for a capacitor to discharge to \(\displaystyle 36.8\%\)of its original value, or have a \(\displaystyle 63.2\%\) drop; therefore, the time constant for this circuit is \(\displaystyle 4ms\).

The time constant, tau, for an RC circuit is the product of the resistance of the resistor and the capacitance of the capacitor:

\(\displaystyle \tau=RC\)

We can use this equation to solve for the resistance of the resistor:

\(\displaystyle R=\frac{\tau }{C}\)

Solving for resistance gives us:

\(\displaystyle R= \frac{4\times 10^{-3}s}{2\times 10^{-6}F}\)

\(\displaystyle R=2000\Omega \:or\:2k\Omega\)

Notice that you could have used the voltage decay equation to solve this problem; however, knowing the definition of time constant simplifies the problem and reduces the amount of calculations.