Response 3 is the correct choice. Electrons will space themselves as far apart as possible, because of charge repulsion; therefore, in a parallel arrangement, they will jump onto each capacitor and “load it up.” The parallel capacitance is calculated by simply adding the individual values. The value would be about 1.75 μF if the three were connected in series, where the formula is (reciprocal of total equals sum of reciprocal of each). The time constant of a circuit is given by the formula τ = RC, and it relates to how fast a capacitor can charge to full capacitance.

First, we need to determine how capacitors C2 and C3 are being added. We can see that they are being added in in series. Remember that capacitors in series are added as reciprocals.

C23 = 0.5μF

Next, we need to determine how we can find the Ceq by simplifying C23 and C1. We can see that Ceq and C1 are in parallel, thus we can directly add the individual capacitances.

Ceq = C23 + C1 = 0.5μF + 0.5μF = 1μF

In order to determine the time, we need to know the total charge stored on the capacitors. Remember that Q = CV, where Q is the total charge, C is the equivalent capacitance, and V is the voltage. We must first find the equivalent capacitance.

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

C23 = 0.5μF

Ceq = C23 + C1 = 0.5μF + 0.5μF = 1μF

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

Q = (1μF)(10V) = 10μC

Additionally, we need to know the current the battery can provide (the charge per unit time). Knowing both the total charge and current will allow us to calculate the time. We can use V = IR to determine the current.

I = V/R = 10V/1Ω = 10A = 10C/sec

We can equate charge and current to determine time.

10μC = 10 C/t

t = 10 C/10 μC = 1 * 106s or 11.6days

In this question, we are asked how the total charge stored on the surface of the capacitors would change if we inserted a dielectric between the parallel plates. As we can see in the equation for capacitance based on physical properties, , where k is the dielectric constant (k = 1 for air; k > 1 for all dielectric materials), ε0 is the constant of the permeability of free space, A is the area of the plates, and d is the distance between the plates.

If we insert a dielectric material, k > 1, the value of C increases, thus the overall capacitance of the circuit increases.

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:

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

Now we can find the charge using the equation .

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.

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.

First, we need to determine how these capacitors are being added. We can see that they are being added in in series. Remember that capacitors in series are added as reciprocals:

Ceq = 0.5μF

Charge is evenly distributed on the surface of the capacitor. If we think back to electric force, we know that positive charges repel other positive charges and negative charges repel other negative charges; thus, the charges are evenly distributed to minimize the force between them. We can see how this looks in diagrammatic form below.