Dielectrics are put between two parallel plates of a capacitor to increase capacitance. This is done by holding voltage constant and increasing charge, or decreasing voltage and holding charge constant.

This can be understood using the equation . Either an increase in charge or a decrease in voltage will increase the capacitance.

First, we need to find the equivalent capacitance for the capacitors in series.

Now we can find the charge using the equation .

The capacitance of any capacitor can be calculated with the formula

Since the capacitors are arranged in parallel, the voltage drop across each will be equal, and in this case 10V. Adding capacitors in parallel is the same as adding resistors in series, where the total capacitance of the circuit is equal to the sum of all individual capacitors.

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.

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.

C₂₃ = 0.5μF

C_eq = C₂₃ + C₁ = 0.5μF + 0.5μF = 1μF

U = ½CV² = ½(1μF)(10V)² = 5 * 10^-5J 

Relevant equations:

The first equation shows that charge  is directly proportional to capacitance , so we want to maximize  in order to find the maximum charge. 

The second equation shows that  is directly proportional to the dielectric constant , and inversely proportional to distance . To increase capacitance, we need  to be large and  to be small.

A smaller distance  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:

Here,  is the voltage after a certain amount of time,  is the initial voltage,  is the time elapsed,  is the resistance of the resistor, and  is the capacitance of the capacitor. We can rearrange this equation in such a way that we solve for capacitance, :

To remove the exponential  from the equation, we need to take the natural logarithm of both sides:

Solving for  gives us:

The time constant, , of an RC circuit is the product of the resistance and the capacitance; therefore, the time constant of this circuit is:

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:

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