Physics 102 Chapters 23 and 24 Notes PDF

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Chapters 23 and 24 Electric Potential and Capacitance 1. The concepts covered in these chapters are: 1. Electric potential 2. Electric potential energy 3. Capacitance 4. Capacitors arranged in series and parallel configuration 5. Air- and dielectric-filled capacitors 6. Potential energy stored in capacitors

2. Potential and Potential Energy. In the previous chapters we learned that charges exert a force on each other according to Coulomb’s law. This means that such forces among the charges can change a given configuration of charges: in the field view of electrostatic interaction the electric fields of various charges interact to make the charges move, i.e., the electric field does work on the charge configuration). A charge configuration (an arrangement of charges) can also be changed by an external agent, i.e. work can be performed by forces external to the charge configuration. This change in charge configuration will show up as a change in the potential energy of the system (the charge configuration).

The change in the potential energy between (or of) two charges is defined as

D Wfield = - D U

…[1a]

D Wext = DU

…[1b]

(These equations are true not only for electrostatic forces but for other conservative forces as well. You used these equations in PHYS-101 for gravitational and elastic forces.) Let’s look at the physical meaning of eq. [1a]. ΔW field = −ΔU = −(U f −Ui ) = Ui −U f U f = Ui − ΔW field or Uf is lower than the initial potential energy Ui.

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Thus, the minus sign in Eq.[1a] means that when you leave charges alone – let the electric field do the work on the charge configuration - they will adjust their configuration in such a way as to lower the potential energy of the configuration. (‘leaving alone’ means you do not add or take away energy from the system of two charges. Such systems are called isolated systems. In isolated systems, since no inflow or outflow of energy is allowed, total mechanical energy is conserved. Thus, if a charge in an isolated system starts moving faster (an increase in kinetic energy), its potential energy must decrease. Other example of isolated system would be a mass+Earth system in which a mass m interacts with the gravitation field of the Earth with no interference from an external agent. Here a falling object will gain kinetic energy and lose an equivalent amount of potential energy. A mass attached to a spring moving on a smooth horizontal surface is another example of an isolated system.)

Similarly, Eq.[1b] means that if you (an external agent) perform positive work on the system the change in the potential energy of the system, DU will be positive, i.e., you will increase the potential energy of the system.

Think about this: if you let two positive charges alone they will move farther away from each other and thereby reduce their potential energy. On the other hand the potential energy of two oppositely charged particles is reduced when they move closer together.

b      F .dr = q E. ∫ ∫ dr b

- DUab = Ua –Ub =

a

…[2]

a

The potential difference DV = Vb - Va is defined as: b    Vb - Va = − ∫ E .dr

…[3]

a

Potential can be defined as the potential energy per unit charge (recall a similar definition of the electric field E as the force per unit charge, qE = F) A point charge Q, creates an electric potential V(r) at a distance r from it given by: V(r) = kQ/r (referenced to infinity, i.e., V (¥) = 0 )

…[4]

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The net potential at a given point due to a collection of charges is simply the algebraic sum of individual potentials n

V = kå i =1

Qi ri

…[5]

Potential energy of two point charges is given by

U12 = kQ1Q2/r

…[6]

We can look at the above equation another way: The charge Q1 creates a potential at r equal to: V1 (r) =!

kQ1 r

Now if you put Q2 at r, Q2 acquires a potential energy: U12 = Q2V1 (r) =!

kQ1Q2 r

In this interpretation Q1 is known as the source charge (source of the electric potential) in whose potential, Q1 acquires a potential energy. Potential energy of a collection or configuration of point charges is the algebraic sum of the potential energy of all the pairs of charges you can form in the configuration. For example, in Solved Example 1 below, Ui is the potential energy of the 3-charge configuration. Here is another example: q

2

r12

q

1

r23 r13

q3

The potential energy of the 3-charge configuration shown above is:

3

U = U12 +U 23 +U 31 =

kq1q2 kq2 q3 kq3q1 + + r23 r31 r12

Potential energy of a single charge in a collection or configuration of point charges is the algebraic sum of the potential energy of all the pairs the given charge forms in the configuration. For example, the potential energy of q1 in the 3-charge configuration above is: U1 = U12 +U 31 =

kq1q2 kq3q1 + r31 r12

Points to remember: 1. Make sure you understand the difference between potential and potential energy. These are similar sounding terms but are different physical quantities. 2. A single charge will create a potential everywhere in space. but you need at least two charges for the potential energy to have any meaning. 3. Potential and potential energy are scalars. (NOTE: The electric potential energy expression looks almost the same as the one for the force between two charges. Be careful! The difference is the potential energy falls off as 1/r, whereas the force falls off as 1/r2.)

4. Eq. [3] can also be expressed as: Ex = - dV/dx (and similar relations in the y- and z-directions, see sec. 23-5 of your text for details). This allows one to calculate the components of E (a vector) by knowing the potential V(x,y,z) which is a scalar.

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Example 1. (How to use Ex = - dV/dx ) In the lecture we derived an expression for the potential at a point at the axis of a charged ring and then calculated the E-field from it. Here is a brief sketch of that derivation. The potential at a point along the axis of a ring of radius a is given by:( see p774 of your textbook)

dq r

a

P

z

V=

∫

kQ kdq kQ = (this is easy since you are dealing with a scalar). Now you can = 2 r r z + a2

obtain the E-field at point P by using Ez = - dV/dz Ez = −

% dV d " kQ % 1 d" =− $ ' = −kQ $ 2 2 ' 2 2 dz dz # z + a & dz # z + a &

Ez = kQ

z (z 2 + a 2 )3/2

…[7]

This expression for E was earlier obtained directly by integration in Example 21-9 on p702 of your text. Example-2. Which potential vs. distance graph describes a constant E-field directed along the positive y-axis?

V

V

V

y [a]

y [b]

y [c]

Answer: [c] 5

3. Capacitance The capacitance C is defined as C = Q/V. C is always positive. When we talk about the charge on a capacitor we mean the magnitude of the charge on one of its plates. The capacitors can be arranged in a series or a parallel arrangement (see sec 24.2 Combination of Capacitors of your textbook for diagrams of capacitors in parallel and series combinations) In a series arrangement: [i] the net capacitance C is given by; 1/C = 1/C1+1/C2+1/C3+….

…[8]

[ii] The magnitude of the charge on every capacitor is the same. [iii] The potential difference across each capacitor is different and is given by Vi = Q/Ci. In a parallel arrangement: [i] The net capacitance C is given by; C = C1 + C2 + C3+….

…[9]

[ii] The potential difference V across each capacitor is the same. [iii] The charge on every capacitor is different and is given by Qi = V Ci. The capacitance of a parallel plate capacitor of plate area A and separation d is given by: C = A e0/d

…[10]

The energy stored in a capacitor is given by U = ½ CV2

…[11]

By using C = Q/V, eq [11] can be cast in terms of other parameters of interest: U = ½ QV or U = ½ Q2/C Insertion of a dielectric medium in a capacitor will change its capacitance by a factor of k, where k is the dielectric constant of the medium, C k = k C0. Prob. 82 was discussed in the lecture.

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4. Solved Examples. 1. Three charges are arranged at the vertices of a rightangle triangle as shown, with Q1= 4 µC =2Q2, and Q3 = 3 µC. [a] Determine the total potential energy of the threecharge system. [b] Determine the work done by an external force to move Q2 from its present position to infinity.

Y Q1

Q3

53o

Q2 X 5.0cm

Solution: [a] The initial potential energy,

U i = 9 ´109 (

-12 ´ 10-12 6´ 10-12 8´ 10- 12 + ) = -2.88J 3 ´10-2 5 ´10-2 4 ´ 10-2

[b] After Q2 has been removed to infinity the potential energy of the remaining charges is given by: -12 ´ 10- 12 ) = - 3.60J U f = 9´ 109 ( 3´ 10- 2 the work done by the external force is given by : D W = DU = Uf - Ui = - 3.60 – ( - 2.88) = - 0.72J Note: in the problem above we have used DW = D U = Uf - Ui , where D U is the change in the potential energy of the charge configuration. We could have also done the problem by looking at it from the point of view of Q2 (in a manner of speaking, charges don’t really have a point of view). In this case D U would be the change in the potential energy of Q2.

D U = Uf(Q2) - Ui(Q2) = 0 – (U12 + U23) = - 9*109(- 1.2 +2.0)10-10 = - 0.72J

2. Consider three charges Q1, Q2, and Q3 placed as shown in the diagram below (disregard Q4 for the moment). [a] Calculate the electric potential energy of the three-charge configuration.

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[b] A fourth particle of mass m = 2.0*10-13kg and charge, Q4= 40.0nC is placed on the xaxis 3.0 cm from Q2 and released from rest. Find the speed of Q4 when it has moved very far away from the three-charge (Q1, Q2, and Q3 ) configuration.

Q1

20.0nC

4.0cm

10.0nC Q2

Q4 3.0cm

40.0nC

4.0cm Q3

- 20.0nC

[a] The total energy of the three charges is U = U12 +U 23 + U13 = U13 = −

9 ×109 × 20 × 20 × 10−18 J = −4.5 ×10−5 J −2 8 ×10

(Why is U12+ U23 = 0 ?) [b] (U + KE)i = (U + KE)f ( conservation of energy from PHYS-101) Ui = KEf (Ui is the initial potential energy of Q4) Ui = U14+ U34+ U24 = U24 = 9*109*400*10-18/3*10-2 = 12.0* 10-5 J = ½*2*10-13v2 (why is U14+ U34 = 0 ?) v = [12]1/2104m/s =3.46*104m/s

3. Consider the two capacitors connected as shown. A potential difference of Vac = 120.0V is maintained between points a and c. The capacitors are fully charged. What’s the potential difference Vab?

2.0 μF

a

6.0μF

b

c

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Solution: Since the capacitors are connected in series, the charges on the two capacitors are equal: q2 = q6 = 2µF ×Vab = 6µF ×Vbc Vab 6 = =3 …[1] Vbc 2 (Note that V’s and C’s in a series arrangement are inversely related.) Vac = Vab +Vbc

…[2]

From [1] and [2]: 120 = Vab + Vab/3 = (4 /3) Vab Vab = 90 V

4. Consider the three capacitors connected as shown. A potential difference of Vac = 120.0V is maintained between points a and c.

4µF [a] What is the potential difference Vbc across the 2µF capacitor? [b] What is the charge on the 4.0µF capacitor? [c] How much energy is stored in the 6.0µF capacitor?

a

b

c 2µF

6µF

Solution: [a] The 4.0µF and the 6.0 µF capacitors can be combined into a single equivalent capacitor of capacitance, 10.0µ F.

10.0 μF

a

2.0μF

b

c

Vab = Q/10.0µ F Vbc= Q/2.0µF or Vab/ Vbc = 2.0/10.0 = 1/5; (Note that if the charges on two capacitors are equal (as in a series circuit) the capacitance and the potential difference across it are inversely related. There is a larger potential difference across the smaller capacitor.) 5Vab= Vbc Also Vab+ Vbc = 120.0V, solving the last two equations gives Vab=20.0V, and Vbc= 100.0V

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[b

Q4 = Vab 4.0*10-6 = 8.0*10-5 C

[c]

U = ½(6.0*10-6)( Vab)2 = 1.2*10-3J

5. Four capacitors, each with a capacitance C = 4.00µF, are connected as shown. The potential difference Vab = 25.0V. [a] What is the equivalent capacitance between a and d ? [b] What is the potential difference between a and d? [c] What is the charge on C4? [d] What is the charge on C1? [e] What is the total electric energy stored in the four-capacitor system?

C1

C2

a d

a

C3 d

b

µF 6µ µF 4µ a b

2.4µ F

b

C4 [a] Cad = (C1C2/ C1+ C2) + C3 = 6.0µF (the resultant of C1 and C2 connected in series is connected in parallel with C3) [b] Vad/Vdb = Cdb/Cad = 4.0/6.0 or Vdb = 1.5Vad Also note that Vad + Vdb = 25.0V

…..[1] …[2]

From eqs. [1] and [2] Vad + 1.5Vad = 25.0V or Vad = 10.0V and Vdb = 15.0V

[c] Q4 = C4* Vdb = 4.0µF* 15.0V = 60.0µC

[d] Q1 = C1* (Vad/2) = 4.0µF* 5.0V = 20.0µC (Note: Vad/2 is the potential across C1)

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[e] U = 0.5C netV ab2 = 0.5*2.4µF* (25.0)2 = 7.5*10-4J

Concepts Questions 1. A spherical balloon contains a positively charged object at its center. (i) As the balloon is inflated to a greater volume while the charged object remains at the center, does the electric potential at the surface of the balloon: (a) increase, (b) decrease, (c) remain the same? (ii) Does the electric flux through the surface of the balloon (a) increase, (b) decrease, (c) remain the same? Answer: (i), (b). The electric potential is inversely proportional to the radius (ii), (c). Because the same number of field lines passes through a closed surface of any shape or size, the electric flux through the surface remains constant.

V = ke

q r

2. (i) In a certain region of space, the electric potential is zero everywhere along the x axis. From this information, we can conclude that the x component of the electric field in this region is: (a) zero, (b) in the +x direction, (c) in the −x direction. (ii) In a certain region of space, the electric field is zero. From this information, we can conclude that the electric potential in this region is: (a) zero, (b) constant, (c) positive, (d) negative.

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Answer: (i), (a). If the potential is constant (zero in this case), its derivative along this direction is zero. (ii), (b). If the electric field is zero, there is no change in the electric potential and it must be constant. This constant value could be zero, but it does not have to be zero. 3. A capacitor stores charge Q at a potential difference V. If the voltage applied by a battery to the capacitor is doubled to 2 V, then: (a) the capacitance falls to half its initial value and the charge remains the same, (b) the capacitance and the charge both fall to half their initial values, (c) the capacitance and the charge both double, (d) the capacitance remains the same and the charge doubles. Answer: (d). The capacitance is a property of the physical system and does not vary with applied voltage. According C =Q/V if the voltage is doubled, the charge is doubled.

4. Two capacitors are identical. They can be connected in series or in parallel. (i) If you want the smallest equivalent capacitance for the combination, (a) do you connect them in series, (b) do you connect them in parallel, or (c) do the combinations have the same capacitance? (ii) Each capacitor is charged to a voltage of 10 V. If you want the largest combined potential difference across the combination, (a) do you connect them in series, (b) do you connect them in parallel, or (c) do the combinations have the same potential difference? Answer: (i), (a). When connecting capacitors in series, the inverse of the capacitances add, resulting in a smaller overall equivalent capacitance. (ii), (a). When capacitors are connected in series, the voltages add, for a total of 20 V in this case. If they are combined in parallel, the voltage across the combination is still 10 V. 5. You have three capacitors and a battery. In which of the following combinations of the three capacitors will the maximum possible energy be stored when the combination is attached to the battery? (a) When in series the maximum amount is stored. (b) When parallel the maximum amount is stored. (c) Both combinations will store the same amount of energy. Answer: (b). For a given voltage, the energy stored in a capacitor is proportional to C: U = C(V)2/2. Therefore, you want to maximize the equivalent capacitance. You do so by connecting the three capacitors in parallel so that the capacitances add. 6. When the potential difference between the plates of a capacitor is doubled, the magnitude of the electric energy stored in the capacitor :

+Q

[a] is doubled remains the same

[b] is halved [d] quadrupled

[c]

7. Three charges are placed on the three corners of an equilateral triangle as shown.. The total potential energy of these charges is: [a] -kQ 2/r

[b] – 1.5kQ2/r

[c] zero [d] + 1.5kQ2/r

8. Two charges are placed along the x-axis as shown. Now, suppose you move the positive charge (while keeping the negative charge at

+Q

-Q X

the origin) to a point 2X away from the origin. As a result of this displacement the electric potential energy of the two charges would: [a] increase

[b] decrease

[c] remain the same

[d] be reduced to zero

9. Two capacitors are connected in series as shown with C1 = 2C2. Let U1 and U2 represent, respectively, the potential energy stored in C1 and C2. Which of the following statements is true? [a] U1 = 2U2

[b] 2U1 = U2

[c] U1 = U2

[c] 4U1 = U2

C1

C2

e=25V 10. Three charges are placed in a line as shown. The potential energy of the three-charge system is: [a] – 2.5kQ2/r

[b] – 1.5kQ2/r

r +Q

[c] zero

[d] + 1.5kQ2/r

r -Q

+Q

13...