PHY136H5 Lecture Notes - Ammeter, Voltmeter, Electric Potential Energy

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8 Apr 2013

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Electric charges tend to move from one position to another of lower electrostatic potential energy in much the same way that masses tend to move to a position of lower gravitational potential energy. Electrostatic potential or simply potential v of a charge q is related to potential energy. A battery is a mechanism such that a constant potential (voltage) difference is maintained across its two terminals. If the terminals are connected through a circuit, charge flows (current) from one terminal to the other. When a conducting element that resists the flow of charge (resistance) is placed in the circuit, charge emerges from the resistor at a different voltage than when it entered the other side. The ratio of the voltage v across the resistor to the current i through it is a constant (for most metallic conductors), independent of i, i. e. (2) is called the resistance of the conductor.

Related Questions

A charge q that moves between two points A and B, and has anelectric force acting on it ( and perhaps several other forces) haswork W done on it by the electric force. The electrostatic force isconservative and so we can define a change in electric potentialenergy, Delta U = -W. We defined electric field using the relationF^^\-> = q E^^\->, and in a similar way we define electricpotential commonly called voltage as
Delta V = V_B - V_A = -W/q=(Delta U)/q or equivalently q Delta V =q (V_B - V_A) = -W=Delta U

A ball has mass 3.0 Âµg and charge -6 nC moves from point A to pointB. It has the same speed at points A and B, and the work done bythe electric field during the move is -1.5 mJ.
(a) What is the change in electric potential energy of the ball?1
(b) What is the change in voltage between points A and B? 2
(c) At which point does the ball have a higher electric potentialenergy?
(d) Which point has the higher voltage?
(e) Has the kinetic energy of the ball changed?

Two parallel plates, each having area A = 2797cm2 areconnected to the terminals of a battery of voltage Vb = 6 V asshown. The plates are separated by a distance d = 0.53cm. You mayassume (contrary to the drawing) that the separation between theplates is small compared to a linear dimension of the plate.

1) What is C, the capacitance of thisparallel plate capacitor?

What is Q, the charge stored on the top plate of thethis capacitor?.

A dielectric having dielectric constant ? = 4.5 is nowinserted in between the plates of the capacitor as shown. Thedielectric has area A = 2797 cm2 and thickness equal to half of theseparation (= 0.265 cm) . What is the charge on the top plate ofthis capacitor?

What is U, the energy stored in thiscapacitor?

The battery is now disconnected from the capacitor andthen the dielectric is withdrawn. What is V, the voltage across thecapacitor?

This set of problems is designed to help you prepare for the lab oncapacitors. Here are some things you should know.
The charge on a capacitor is proportional to the voltage. Theconstant of proportionality is the capacitance:
q=VC.
The simplest type of capacitor is a parallel plate capacitorconsisting of twoparallel conducting plates...DUH. If the plateseach have area A and
are separated by a distance d, the capacitance will be
C=Ke0A/d,
where e0 is the permittivity of free space (you measured that lastweek) and K is the dielectric constant of what ever material isbetween the plates. If the space between the plates is filled withvacuum, the dielectric constant is exactly 1. Air is pretty darnclose to 1.
When a capacitor is charged in a series circuit with a resistor,the charge builds up with time. Consider the circuit below.

When the switch, S, is thrown to position A, the battery, withvoltage V0, will supply a current through the resistor, R, thatcharges the capacitor C. The voltage across the capacitor at anygiven time is given by
V=V0(1-e-t/RC).
The product RC is known as the time constant. After waiting a timeequal to 1 time constant, the voltage on the capacitor is(1-e-1)=0.632 of V0. After 5RC the capacitor is nearly fullycharged and has a voltage of 0.993V0. Eventually, the voltagebuilds up to V0.
When the switch is thrown to position B, the capacitor willdischarge through the resistor. The voltage as a fuction of timewill be given by
V=V0(e-t/RC).
After a long enough time, the capacitor will fully discharge theresistor.

1. [1pt]
A parallel plate capacitor has plates with length and width of 5.7cm and are separated by 7.0 mm. The capacitance is measured to be1.479Ã10-11 F. What is the dielectric constant of the materialbetween the plates?

Correct, computer gets: 3.60

2. [1pt]
Consider a series circuit like the one shown above with a resistorof R=80. kilo-ohms and a capacitor of C=8.0 uF. What is the timeconstant of the circuit?

Correct, computer gets: 0.64 s

Hint: Don't forget to convert to Farads and ohms from uF andkilo-ohms.
3. [1pt]
If the battery has a voltage of 12 V, what is the voltage acrossthe capacitor after the switch has been in position A for 2 timeconstants?

Correct, computer gets: 10.38 V
4. [1pt]
What is the final charge on the capacitor after a long time?

Correct, computer gets: 9.60E-05 C
5. [1pt]
Now let's say that the capacitor has been fully charged and wethrow the switch to position B. How long will it take for thevoltage on the capacitor to be 0.891 V?

6. [1pt] What is the charge on the capacitor at this time?

Answer: ____________

PHY136H5 Lecture Notes - Ammeter, Voltmeter, Electric Potential Energy

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