Lecture 7 Potential from a distribution of charges 1 q V = i 4⇡✏ r 0 i i § Smooth distributionX
1 q 1 ⇢ V = i = dV 4⇡✏ r 4⇡✏ r 0 i i 0 X Z § Calculating the electric potential from a group of point charges is usually much simpler than calculating the electric field • It’s a scalar Electric Potential from Two Oppositely Charged Point Charges § The electric field lines from two oppositely charge point charges are a little more complicated § The electric field lines originate on the positive charge and terminate on the negative charge § The equipotential lines are always perpendicular to the electric field lines § The red lines represent positive electric potential § The blue lines represent negative electric potential § Close to each charge, the equipotential lines resemble those from a point charge
22 Electric Potential from Two Identical Point Charges § The electric field lines from two identical point charges are also complicated § The electric field lines originate on the positive charge and terminate at infinity § Again, the equipotential lines are always perpendicular to the electric field lines § There are only positive potentials § Close to each charge, the equipotential lines resemble those from
a point charge 23 Example: Superposition of Electric Potential (1) § Assume we have a system of three point charges:
q1 = +1.50 µC
q2 = +2.50 µC
q3 = -3.50 µC
§ q1 is located at (0,a)
q2 is located at (0,0)
q3 is located at (b,0) a = 8.00 m and b = 6.00 m Question: What is the electric potential at point P located at (b,a)?
30 Example: Superposition of Electric Potential (2) r1 Answer: § The electric potential at point P is r2 r given by the sum of the electric 3 potential from the three charges
31 Conductor in E-field
§ Charges can move freely along conductor boundary. § Recall: electrostatic shielding: E=0 inside § Surface of a conductor is an equipotential surface. § E-field is perpendicular to conductor’s surface
7 Electric Potential Energy for a System of Particles
§ So far, we have discussed the electric potential energy of a point charge in a fixed electric field § Now we introduce the concept of the electric potential energy of a system of point charges § In the case of a fixed electric field, the point charge itself did not affect the electric field that did work on the charge § Now we consider a system of point charges that produce the electric potential themselves § We begin with a system of charges that are infinitely far apart • This is the reference state, U = 0 § To bring these charges into proximity with each other, we must do work on the charges, which changes the electric potential energy of the system
37 Electric Potential Energy for a Pair of Particles (1) § To illustrate the concept of the electric potential energy of a system of particles we calculate the electric potential energy of a system of two
point charges, q1 and q2 . § We start our calculation with the two charges at infinity
§ We then bring in point charge q1 • Because there is no electric field and no corresponding electric force, this action requires no work to be done on the charge
§ Keeping this charge (q1) stationary, we bring the second point charge (q2)
in from infinity to a distance r from q1
• That requires work q2V1(r)
38 Electric Potential Energy for a Pair of Particles (2) § So, the electric potential energy of this two charge system is
where § Hence the electric potential of the two charge system is
§ If the two point charges have the same sign, then we must do positive work on the particles to bring them together from infinity (i.e., we must put energy into the system) § If the two charges have opposite signs, we must do negative work on the system to bring them together from infinity (i.e., energy is released from the system)
39 Electric Potential Energy Consider three point charges at fixed positions. Question: What is the electric potential energy U of the assembly of these charges?
d12 q1 q2
d13 d23
q3
11 Example: Electric Potential Energy (2)
Answer: § The potential energy is equal to the work we must do to assemble the system, bringing in each charge from an infinite distance
§ Let’s build the system by bringing the charges in from infinity, one at a time
41 Example: Electric Potential Energy (3)
§ Bringing in q1 doesn’t cost any work
§ With q1 in place, bring in q2
12 § We then bring in q3.The work we must do to bring
q3 to its place relative to q1 and q2 is then: q q U = U + U = kq 1 + 2 13 23 3 d d ✓ 13 23 ◆ q q q q q q U = U + U + U = k 1 2 + 1 3 + 2 3 tot 12 13 23 d d d ✓ 12 13 23 ◆ A sum of pair interaction energy.
42 Alternative way
Let’s calculate energy of a particle in a total potential created by all other particles. And then sum over all particles. q Potential created by 2 &3 at 2 q3 V1 = k + position of 1 is d d ✓ 12 13 ◆ q q Potential created by 1 &3 at 1 3 V2 = k ++ position of 2 is d d ✓ 12 23 ◆ q q Potential created by 1 &2 at V = k 1 ++ 3 position of 3 is 3 d d ✓ 13 23 ◆
42 § Total energy: Utot = q1V1 + q2V2 + q3V3(?)
Utot = q1V1 + q2V2 + q3V3 = q2 q3 q1k + + ???2??? d12 d13 ✓ ◆ I counted each interaction twice! q1 q3 q2k + + d12 d23 q1V1 + q2V2 + q3V3 ✓ ◆ Utot = q1 q3 2 q3k + 1 d13 d23 U = q V ✓ ◆ tot 2 i i q q q q q q N =2k 1 2 + 1 3 + 2 3 X d d d ✓ 12 13 23 ◆ Capacitors
§ Capacitors are devices that store energy in an electric field. § Capacitors are used in many every-day applications • Heart defibrillators • Camera flash units
§ Capacitors are an essential part of electronics. • Capacitors can be micro-sized on computer chips or super-sized for high power circuits such as FM radio transmitters.
§ Capacitors come in a variety of sizes and shapes.
The world's first integrated circuit included one capacitor Capacitor controlling power which enters Fermilab, a particle accelerator laboratory in Batavia, IL
2 Capacitance § Concept: A capacitor consists of two separated conductors, usually called plates, even if these conductors are not simple planes. § We will define a simple geometry and generalize from there.
§ We will start with a capacitor consisting of two parallel conducting plates, each with area A separated by a distance d.
§ We assume that these plates are in vacuum (air is very close to a vacuum).
3 Parallel Plate Capacitor (1)
+ - + + - - + + - -
+ - § We charge the capacitor by placing • a charge +q on the top plate • a charge -q on the bottom plate § Because the plates are conductors, the charge will distribute itself evenly over the surface of the conducting plates. § The electric potential, V, is proportional to the amount of charge on the plates. V = dE dq ⇥ q E = = V = 0 0A 0A
4 Parallel Plate Capacitor (2)
§ The proportionality constant between the charge q and the electric potential difference V is the capacitance C. § We will call the electric potential difference V the “potential” or the “voltage” across the plates. § The capacitance of a device depends on the area of the plates and the distance between the plates, but does not depend on the voltage across the plates or the charge on the plates. § The capacitance of a device tells us how much charge is required to produce a given voltage across the plates.
5 Definition of Capacitance
§ The definition of capacitance is
§ The units of capacitance are coulombs per volt. § The unit of capacitance has been given the name Farad (abbreviated F) named after British physicist Michael Faraday (1791-1867)
§ A farad is a very large capacitance • Typically we deal with µF (10-6 F), nF (10-9 F), or pF (10-12 F) © Pixtal/age Fotostock RF
6 Charging/Discharging a Capacitor (1)
§ We can charge a capacitor by connecting the capacitor to a battery or to a DC power supply. § A battery or DC power supply is designed to supply charge at a given voltage. § When we connect a capacitor to a battery, charge flows from the battery until the capacitor is fully charged. § If we then disconnect the battery or power supply, the capacitor will retain its charge and voltage. § A real-life capacitor will leak charge slowly, but here we will assume ideal capacitors that hold their charge and voltage indefinitely.
7 Parallel Plate Capacitor (1)
§ Consider two parallel conducting plates separated by a distance d
§ This arrangement is called a parallel plate capacitor. § The upper plate has +q and the lower plate has –q. § The electric field between the plates points from the positively charged plate to the negatively charged plate. § We will assume ideal parallel plate capacitors in which the electric field is constant between the plates and zero elsewhere. § Real-life capacitors have fringe field near the edges.
9 Parallel Plate Capacitor (2)
We can calculate the electric field between the plates using Gauss’ Law
10 Parallel Plate Capacitor (3)
§ Now we calculate the electric potential across the plates of the capacitor in terms of the electric field. § We define the electric potential across the capacitor to be V. § We carry out the integral in the direction of the blue arrow.
11 Parallel Plate Capacitor (4)
Remember the definition of capacitance…
… so the capacitance of a parallel plate capacitor is Variables: A is the area of each plate d is the distance between the plates
Note that the capacitance depends only on the geometrical factors and not on the amount of charge or the voltage across the capacitor.
12 Example: Capacitance of a Parallel Plate
§ We have a parallel plate capacitor constructed of two parallel plates, each with area 625 cm2 separated by a distance of 1.00 mm. Question: What is the capacitance of this parallel plate capacitor?
Answer: A parallel plate capacitor constructed out of square conducting plates 25 cm x 25 cm separated by 1 mm has a capacitance of about 0.5 nF.
13 Example: Capacitance, Charge, and Electrons … Question: A storage capacitor on a random access memory (RAM) chip has a capacitance of 55 nF. If the capacitor is charged to 5.3 V, how many excess electrons are on the negative plate?
Answer: Idea: We can find the number of excess electrons on the negative plate if we know the total charge q on the plate. Then, the number of electrons n=q/e, where e is the electron charge in coulomb. Second idea: The charge q of the plate is related to the voltage V to which the capacitor is charged: q=CV.
15 Capacitors can be dangerous!
§ 1 eV ~ 10000K § 100 eV ~ 100000K!
29 Cylindrical Capacitor (1)
§ Consider a capacitor constructed of two collinear conducting cylinders of length L • Example: coax cable
§ The inner cylinder has radius r1 and
the outer cylinder has radius r2 § Both cylinders have charge per unit length λ with the inner cylinder having positive charge and the outer cylinder having negative charge
16 Cylindrical Capacitor (2)
§ We apply Gauss’ Law to get the electric field between the two cylinders using a Gaussian surface with radius r and length L as illustrated by the red lines
§ … which we can rewrite to get an expression for the electric field between the two cylinders
r1< r < r2
17 Cylindrical Capacitor (3)
§ As we did for the parallel plate capacitor, we define the voltage
difference across the two cylinders to be V = |V1 – V2|
§ Thus, the capacitance of a cylindrical capacitor is
Note that C depends on geometrical factors only.
18 Spherical Capacitor (1)
§ Consider a spherical capacitor formed by two concentric
conducting spheres with radii r1 and r2
19 Spherical Capacitor (2)
§ Let’s assume that the inner sphere has charge +q and the outer sphere has charge –q § The electric field is perpendicular to the surface of both spheres and points radially outward
20 Spherical Capacitor (3) § To calculate the electric field, we use a Gaussian surface
consisting of a concentric sphere of radius r such that r1 < r < r2
§ The electric field is always perpendicular to the Gaussian surface so
…makes sense! § … which reduces to
21 Spherical Capacitor (4)
§ To get the electric potential we follow a method similar to the one we used for the cylindrical capacitor and integrate from the negatively charged sphere to the positively charged sphere
§ Using the definition of capacitance we find
§ The capacitance of a spherical capacitor is then
22 Capacitance of an Isolated Sphere
§ We obtain the capacitance of a single conducting sphere by taking our result for a spherical capacitor and moving the outer spherical conductor infinitely far away § Using our result for a spherical capacitor…
§ …with r2 = ∞ and r1 = R we find
…meaning V = q/4πε0R (we already knew that!)
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