### SAT Physics Electric Potential - Motion Of Charges And Potential

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** SAT Physics Electric Potential - Motion Of Charges And Potential**

**MOTION OF CHARGES AND POTENTIAL**

When charges are released in electric fields, the charges experience a force, causing them to accelerate parallel to electric field vectors. Positive charges accelerate in the direction of the electric field. Negative charges move opposite the electric field. While in motion, the charges experience a change in potential, known as a potential difference. If the charges change speed, then work is done on or by the charges. Changes in speed are associated with changes in kinetic energy, which result from changes in potential energy. Energy is conserved in these processes.

**Potential Difference of Moving Charges**

When a charge is released in an electric field, the charge moves parallel to electric field lines, causing a change in distance, Δd (uniform field) or Δr (field of a point charge). The change in distance results in a change in potential, ΔV; specifically known as a potential difference.

ΔV= V

In addition, the term “potential difference” can refer to the difference in potential between separated charges, such as the potential between two charged plates._{f}- V_{t}**Work of Electricity**

When a charge accelerates, it changes speed and experiences a change in kinetic energy, ΔK The work-kinetic energy theorem states that changes in kinetic energy are equal to work, W= ΔK If kinetic energy is changing, another energy in the system must be experiencing an equal but opposite change in order to ensure that energy is conserved. In problems involving a moving charge, the change in kinetic energy is offset by an equal change in electric potential energy. This is directly proportional to the potential difference through which the charge moves.

W

Note that the above equation equates the values to one another but does not indicate the correct sign on each value. When a charge speeds up, the change in kinetic energy and work are both positive. In order for kinetic energy to increase, the electric potential energy must decrease. This means the sign on the change in electric potential energy is the opposite of the sign on the change in kinetic energy and work. When a charge slows down, the change in kinetic energy and work are negative while the sign on the charge in electric potential energy is positive._{E}= ΔK= -ΔU_{E}= -qΔV= -q(V_{f}- V_{i})Solving for the correct sign on work using the change in electric potential energy equation, W= -ΔU

_{E}= -(qΔV), is complicated since the equation contains a negative sign and all the variables may be either positive or negative. Most physics problems in beginning courses involve situations where the work of the electric force is positive. As a result, solving for the absolute value of work is actually easier.

**Work of Electricity**

**WHAT'S THE TRICK?**

Work is equal to a change in energy. This problem involves equipotential lines. The energy that is changing is electric potential energy, A This is related to the potential difference ΔU

_{E}, This is related to the potential difference,ΔV,through which the 2-coulomb charge moves.

W = -ΔU

W = -ΔU

W = +12 J

Work involves change, and the sign is often important. Be very careful to subtract the initial potential difference from the final potential difference and to include the given signs. Note that the positive 2-coulomb charge will be attracted to the negative 2 V equipotential line and will therefore speed up. An increase in speed is positive for both work and change in kinetic energy._{E}= -q ΔV = -q(V_{f}- V_{i})W = -ΔU

_{E}=-q ΔV = -(2 C)[(-2 V) - (+4 V)]W = +12 J

**Conservation of Energy**

Conservation of energy dictates that the total energy (sum of the kinetic and potential energies) of a closed system must have the same value at any two points during a process.

K

Electricity problems involve electric potential energy, U_{1}+U_{1}= K_{2}+U_{2}_{E}= qV.

_{}

**Conservation of Energy**

**WHAT'S THE TRICK?**

Remember conservation of energy. When the electron is located at the negative plate, it has only potential energy. When it reaches the positive plate, the electron has reached maximum speed. So all of the potential energy has been entirely converted to kinetic energy.

**CAPACITORS**

Charged plates are able to store excess charge and electric energy. This property has a useful application for electrical circuits. The capacity of charged plates to hold excess charges is an important factor. In fact, charged plates are known as

**capacitors.**

**Capacitance**

**Capacitance**, C, which is measured in farads (F), is essentially the ability of a capacitor to store charge and energy. Two oppositely charged parallel plates are shown in Figure 11.5.

**Figure 11.5**Oppositely charged pallel plates

The capacitance of the plates is directly proportional to the area, A, of one of the plates and inversely proportional to the distance of the plate separation, d.

_{0}= 8.85 x 10

^{-12}C

^{2}/N • m

^{2}, is needed to turn the proportionality into an equation solving for capacitance, C. Multiplying by a value such as this constant would be difficult without a calculator. Instead, the SAT Subject Test in Physics will most likely focus on factors affecting capacitance, as demonstrated in Example 11.5.

**Charging a Capacitor**

A battery or a power supply is needed to provide the electric potential needed to charge capacitors or to run circuits. Batteries, like capacitors, are made up of plates of conducting material. The main difference is the presence of chemicals in a battery, which undergo a continuous reaction to keep the plates of the battery loaded with a fixed charge. The fixed charge on the plates of the battery creates a constant potential difference (voltage) across the plates and terminals (ends) of the battery. This potential provides the electrical pressure needed to charge capacitors and/or push charges through circuits. A power supply is a device that functions like a battery. It is plugged into an electrical outlet and adjusts the voltage delivered by the power company to a desired level.

Figure 11.6 shows two versions of the same capacitor connected to a 6-volt battery and uses circuit symbols for the battery and the capacitor.

**Figure 11.6. A capacitor connected to a battery**

In Figure 11.6(a), the switch is open and the capacitor is initially uncharged, Q = 0. An uncharged capacitor has no potential, V

_{c}= 0. In Figure 11.6(b), the switch has been closed for a long time, allowing the potential of the battery to push charges onto the capacitor. As the capacitor fills with charge, a potential (pressure) builds up on the plates of the capacitor. The capacitor will continue to fill as long as the potential of the battery is greater than the potential of the capacitor. This process happens quickly at first. As more and more charge builds up on the capacitor, though, forcing additional charges on the capacitor becomes more difficult. Eventually, the potentials become equal (V

_{c}= V

_{batt}), charging stops, and the capacitor is full. The amount of charge, Q, stored on a capacitor is a function of its capacitance, C, and its potential, V.

Q = CV

**Energy of a Capacitor**

Capacitors also store energy. Once a capacitor has been charged, the charges will remain on its plates as long as no pathway is provided for the charges to move from one plate to the other. The battery can even be removed from the circuit and the charges will remain in place on the capacitor. The charges are essentially held in a position to be used later. Therefore, the energy of a capacitor is potential energy, U

_{c}.

**Capacitors**

**(A)**Determine the effect on the charge stored on the capacitor if the potential of the power supply is cut in half.

**WHAT'S THE TRICK?**

The charge on a capacitor is tied to its capacitance, C, and the potential, V, across its plates. When a capacitor is fully charged, its potential will be equal to the potential of the power supply. Capacitance depends on the area, A, of the plates and the distance, d, between them. These quantities are not mentioned in the problem and are therefore assumed to remain constant. When capacitance is constant, the charge, Q, stored on the capacitor is directly proportional to the potential, V, of the capacitor. This potential is being cut in half

**(B)**How does this change in power supply and potential affect the energy stored on the capacitor?

**WHAT'S THE TRICK?**

Care must be taken in analyzing energy. The first equation that comes to mind is

**Discharging a Capacitor**

A full capacitor stores both charge and energy. If a wire or a circuit is connected between the terminals of the capacitor, the potential difference and stored energy will cause charges to move from one plate of the capacitor to the other. This process occurs very rapidly at first and tapers off as the potential of the capacitor approaches zero. Eventually, the plates of the capacitor will reach a neutral charge.

One application involving capacitors is flash photography. A battery cannot deliver the surge of charge needed to create a quick and bright source of light. Instead, the battery is used to charge a capacitor slowly. Once the capacitor has been fully charged, a flash photo can be taken. Depressing the button on the camera to take the photo also closes a circuit between the plates of the capacitor. The charges stored on the capacitor surge from one plate of the capacitor to the other. Along the way, the charges pass through a specialized lightbulb, creating a quick and blinding flash.

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