Electric Potential and Potential Energy

Electric Potential Work-energy theorem: Change in potential energy = work done m Higher PE Gravitational Potential Energy It requires a certain amount of work to raise an object • Chapter 17 (Giancoli) of mass m from the ground to some distance above the • All sections except 17.6 (electric m Lower PE ground. dipoles) i.e. We have increased the potential energy of the object.

PE=W=Force x Displacement

Electric Potential Energy ⎛ kqQ ⎞ Find the work done in bringing a charge q from infinitely ∆ W = − F∆ r = −⎜ 2 ⎟⋅∆ r far away to a distance R from charge Q. ⎝ r ⎠ Q R R ∆r kqQ PE = ∆W = - ∆r F ∫ ∫ r 2 + +q ∞ ∞ R R ⎡1⎤ Charge is moved towards R = kqQ⎢ ⎥ by increments of ∆r ⎣ r ⎦∞

kqQ For a small displacement ∆r, the work done is: PE = *Note: PE 0 when R ∞ R ∆W = force x displacement = -F • ∆r

(we have a negative sign as the direction of the force is This is the PE of a charge q when it is a distance R from Q. opposite to the direction of the displacement)

• PE is a scalar quantity If the PE is negative (when the charges have opposite signs), then the work is done by the charge, decreasing its PE.

• The sign of the charges must be kept in all calculations Displacement

+ F q • Depending on the signs of Q and q, the PE could be positive or negative

If the PE is positive (when the charges are both positive or both negative), then work must be done on the charge q to bring it closer to Q, increasing its PE. Displacement F q

© Z. Altounian 1 Electric Potential For a point charge, Q´ ⎛ kqQ′ ⎞ 1 V = ⎜ ⎟ Definition: Define the electric potential, V, at a point by the ⎝ r ⎠ q PE of a charge at that point divided by the value of the charge. kQ′ V = r PE V = Units for V: J/C q This is the potential a distance r from the charge Q´. * This property of the charge Q´ is more fundamental than PE. The potential, similar to PE, is a scalar quantity. i.e. We must keep the sign of the charges in all calculations If we know the potential at a point in space, when we place a charge Q at that point, its PE can be directly give by:

PE = QV

The potential V at a point can be due to the presence of a Example : a) Calculate the potential at point P in the figure; single charge or due to many charges.

Q1 = 3 µC Q2 = –6 µC For many charges, Q1, Q2, Q3, ... , the potential at a point P is P given by the superposition principle: • + r1=6 cm r2=8 cm P Vp = V1 + V2 + V3 +… r1 Q1 r3 kQ kQ kQ r2 V = V + V = 1 + 2 + 3 +… P 1 2 r1 r2 r3 Q Q kQ kQ 2 3 = 1 + 2 r1 r2 * Remember to include the sign of the charges Q1, Q2, Q3, ... , (9×109 N⋅m2 /C2)(3×10−6 C) (9×109 N⋅m2 /C2 )(−6×10−6 C) when calculating Vp. = + (0.06m) (0.08m) * If a charge Q is placed at point P its potential energy will be: 5 PE = Q Vp = −2.25×10 J /C

b) What is the potential energy of a charge Q3=2 µC when Example: Along a line joining the two charges, Q1 = 40 µC placed at P? and Q2 = -60 µC, separated by 4 m, where does V vanish?

Q1 = 3 µCQ3= 2 µC Q2 = –6 µC

• + r1=6 cm + r2=8 cm P PE = Q VP Q1 = 40 µC Q2 = –60 µC = (2 × 10−6C)(−2.25 × 105J / C) P = −0.45 J • + r1= x r2= 4-x Potential is not affected by the presence of a charge at that point. (Charges cannot contribute to their own potential) Q2 has more charge than Q1, therefore P must be closer to Q1 If free to move, an object will always try to lower its (V ∼ Q/r) PE.

40×10−6 N ⋅m2 / C 60×10−6 N ⋅m2 /C Are there other solutions? ⇒ − = 0 − (x)m (4 x)m To the left of Q1 ⇒ 40(4 − x) = 60x Q1 = 40 µC Q2 = –60 µC P Solving for x, we get x = 1.6 m + r 1= x

r2= x + 4

Vp = V1 + V2 = 0

Q = 40 µC Q = –60 µC kQ1 kQ2 1 2 + = 0 P r1 r2 + −6 2 −6 2 r = 8 m × ⋅ × ⋅ 1 40 10 N m /C 60 10 N m /C r =12 m ⇒ − = 0 2 (x)m (x + 4)m Also, V= 0 at +ve infinity, and -ve infinity ⇒ + = Therefore, the potential energy of a charge Q at P is equal to 40(x 4) 60x PE (∞) = 0.

Solving for x, we get x = 8 m

Sometimes it is not necessary to know the absolute value of V Example or PE. m h Example For convenience, we m can take the PE of table the object of mass m h to be zero when it is on the ground. ground ground Then, at a height h above the ground, its PE, relative to the In this example, we may choose to take the zero of PE at the ground is mgh. top of the table. Then the PE of the object a distance h above the table is mgh.

* The zero of PE is chosen for convenience.

© Z. Altounian 3 Similarly for the electric potential and potential energy, Experimentally, we always perform measurements in the sometimes we are interested in the changes in V and PE. laboratory (and not at infinity!), so the symbol ‘∆’ is not really used, i.e. V ≡∆V. ∆ PE ∆ V = q Historically, the difference in potential was measured or calculated with the unit of volts [V] while the absolute potential was measured or calculated with the unit of joules Where ∆V is the potential difference and ∆PE is the change in per coulomb [J/C]. The two units are equivalent and are now the potential energy, measured from a convenient reference used interchangeably. point. 1V = 1 J/C

Potential Energy of a System of Charges : Example: What is the PE of the two charges Q1 and Q2 at A and B? A B

The potential energy (PE) of a charge or a collection of Q1 Q2 charges is the amount of work needed in assembling the R charges. Bring one charge at a time from ∞, calculating the work done in each case. The PE of the system of the two charges is the total work done.

For the first charge, Q1, the work done is zero, as VA is zero, W1 = 0.

For the second charge, Q2, the potential at point B is not zero. kQ 1 ⎛ kQ1 ⎞ kQ1Q2 VB = W2 = VBQ2 = ⎜ ⎟Q2 = R ⎝ R ⎠ R kQ Q PE = W + W = 1 2 1 2 R

Note: This is also the PE of Q1 or Q2.

Example : Consider three charges, Q1 = 2 µC, Q2 = – 1 µC 1. Start with Q1. The work done, W1, in bringing the charge and , Q3 = 3 µC located at points A, B and C as shown in the diagram. Calculate the PE of the three charges. from ∞ to point A is given by:

W1 = Q1 VA = 0 {as VA = 0 (as no charges are present )}

AB C 2. Next bring Q2. The work done is W2 = Q2 VB. VB is not 2 m 3 m zero as Q1 is present at point A.

Q1 Q2 Q3 AB 2 m

Solution : Assume that the three charges are infinitely far Q1 Q2 away. Bring the charges, one at a time, each time calculating the work done on the charge. The total work done is the PE of W = Q V = Q (kQ /R) = – 0.009 J the system of the three charges (work-energy theorem). 2 2 B 2 1

© Z. Altounian 4 3. For the last charge, Q3, the work done is W3 = Q3 VC, Example : Two charges, Q1=40 µC and Q2= - 60 µC are 4 m where both Q1 and Q2 contribute to the potential at point C. apart. How much work is done in moving a charge Q = 50 µC from point A to point B?

Q = 40 µC Q = -60 µC AB C 1 2 2 mA 2 m 2 m 3 m + • Q1 Q2 Q3 2 m

W3 = Q3 { (kQ1/R1) + (kQ2/R2) } = 0.0018 J B

Total work done = W1 + W2 + W3 = – 0.0072 J Work done = change in potential energy ∴ PE = – 0.0072 J = QVB –QVA

Work done = Q ( VB -VA) Q1 = 40µC Q2 = -60µC 2mA 2m + • = 50 x10-6 C(-6.39 x 104 J/C - (-9 x 104 J/C)) 2m r =2 2m = 1.3 J r1 =2 2m 2 B

kQ1 kQ2 VB = + r1 r2 (9×109 N ⋅m2 /C2 )(40×10−6 C) (9×109 N ⋅m2 /C)(60×10−6 C) = − (2 2) m (2 2) m = −6.39×104 J /C

Conservation of Energy Example : Two fixed charges, Q 1= 6 µC and Q2 = 4 µC are 4 m apart. A charge of 2 µC and mass 2 g is placed at A. If In many situations, it is very useful to use the conservation of this charge is released from rest where will it go and can one energy principle. determine its final velocity (if yes, find the final velocity)?

[Total Energy at point A] = [Total Energy at point B]

(KE + PE) = (KE + PE) A B Q1 = 6 µC Q2 = 4 µC Q = 2 µC + 4 m + + 1 1 2 m mv2 + QV = mv2 + QV A 2 A A 2 B B Fixed charges

© Z. Altounian 5 The charge Q will move in the direction of the force acting The charge will move in the direction of the force acting on it. on it. Therefore, the charge Q moves to the right. The charge moves to the right.

Q1 = 6 µC Q2 = 4 µC Q= 2 µC Q1 = 6 µC Q2 = 4 µC + 4 m + + F + 4 m + • + F 2 m 2 m A A

The charge moves to the right and reaches infinity.

(P.E. + K.E.)A = (P.E. + K.E.)∞ QV = 1 mv2 00 A 2 ∞ 1 2 1 2 QVA + mvA = QV∞ + mv∞ 2 2 −6 4 1 −3 2 (2×10 C)(2.7×10 J/C) = (2×10 kg)(v∞ ) Therefore, 2 QV = 1 mv2 Need to evaluate V A 2 ∞ A

kQ1 kQ2 VA = V1 + V2 = + r1 r2 v∞ = 7.35 m/s

(9 × 109 Nm2/ C2)(6 × 10−6 C) (9 × 109 Nm2/ C2)(4 × 10−6 C) = + (4 + 2) m (2) m

= 2.7×104 J / C

Example : Two charges, Q1= -50 µC and Q2=10 µC are 3 m apart. A particle of charge Q =1 µC and mass 5 g is placed at Q1 = -50 µC Q2 = 10 µC Q =1µC 3m E P E point P, 1 m to the right of Q as shown. The particle is 1 2 2 + released from rest. 1m a) What is the force acting on the particle? kQ 9 2 −6 b) Can one determine the “final” velocity of the particle. 1 (9×10 N ⋅m / C)(50×10 C) 4 E1 = 2 = 2 = 2.81×10 N / C r1 (3m +1m)

Q1 = -50 µC Q2 = 10 µC Q = 1 µC 3m P kQ (9×109 N ⋅m2 / C)(10×10−6 C) E = 2 = = 8.99×104 N / C + 1m 2 2 2 r2 (1m) a) Find the resultant E - field : (F = QE) The magnitude of E : p E p = E2 − E1 E = E + E p 1 2 = 6.18×104 N / C

© Z. Altounian 6 The direction of Ep: b) It seems that the particle will reach infinity. Using conservation of energy: Q1 = -50 µC Q2 = 10 µC Q =1µC 3m E P E + 1 2 (PE + KE)p = (PE + KE)∞ 1m E 0 0 p 1 2 1 2 F QVp + 2 mv p = QV∞ + 2 mv∞ The magnitude of F: 1 2 QVp = mv∞ F = Q E p 2 − 6 4 kQ kQ = (1 × 10 C )( 6.18 × 10 N / C ) V = V + V = 1 + 2 p 1 2 r r = 0.0618 N 1 2 (9 × 109Nm2 / C2 )(−50 × 10−6C) (9 × 109 Nm2 / C2 )(10 × 10−6C) = + The direction of F: Same as for Ep (4)m (1)m The particle of charge Q moves to the right = −2.25×104V

2QVP Q = -50 µC Q = 10 µC Q =1µC v = 1 2 ∞ m 3m + PP´´P′ 2(1 × 10−6C )( −2.25 × 104V ) F F ⇒ v = F=0 ∞ − 3 ( 5 × 10 )kg As the particle moves to the right, the net force acting on it starts to decrease until it reaches a point (P´) where this force becomes zero and changes its sign (direction). The velocity has an imaginary value!!! The particle cannot reach ∞ As a result, the particle starts to slow down, and at a certain point (P´´) it will stop (v=0) and reverse its direction of motion i.e. return. Where is point P´ ?

At point, P′, the force on the particle becomes zero and then Where will the particle stop and start to return? (i.e. where is changes its direction. point P´´ ?)

F = 0 when the E-field is zero. Q1 = -50 µC Q2 = 10 µC Q = 1µC p P´´ ∴ E = E 3m 1 2 + 1m x Q1 = -50 µC Q2 = 10 µC Q =1µC 3m E P′ E When velocity = 0; apply the conservation of energy: + 1 2 x (P.E. + K.E.)p = (P.E. + K.E.) kQ kQ (50×10−6 C) (10×10−6 C) p′′ 1 2 ⇒ = 2 = 2 2 2 00 (3m + x) x 1 2 1 2 r1 r2 QV + mv = QV + mv p 2 p p′′ 2 p′′ (3m + x) 2 ⇒ = 5 Solving for x, we get x = 2.43 m x 2

Due to chemical reactions, positive charges accumulate on the kQ kQ ⇒ 1 + 2 = −2.25 × 104 V positive terminal (anode) and negative charges accumulate on (3 m + x) x the negative terminal (cathode).

_ Symbol for the battery + Solving for x; x = 12 m anode cathode

or simply

Electric Field Lines Electric field lines of single charges

• A graphical representation of electric fields.

• Direction: E-field lines start from a positive charge and end on a negative charge.

• Magnitude: The density of the electric field lines per unit area is proportional to the magnitude of the E-field.

A week and a stronger electric field Electric field lines for two equal charges

Charges have the same sign Charges have opposite sign

Charge = – Q Charge = – 2Q (Electric Dipole)

(Week field) (Strong field)

© Z. Altounian 8 Field inside two oppositely charged parallel plates Uniform E-fields If a pair of conducting parallel plates are connected to the terminals of a battery, charges are deposited on the plates. _ _ + + _ The electric field lines + + _ + _ are straight. _ + _ + E + _ + _ + _ _ + E _ + + _ + _ The E-field is uniform _ + _ + + _ A Uniform E-field exists between the two plates.

Work done in moving a positive charge q from the negative W = change in PE = qV plate to the positive plate is: d + – ∴ qEd = qV Work done = Force x displacement + – + E – + – E = V/d For a small displacement, ∆r, + + – + – ∆W = F ∆r Alternate unit for E: [V/m] instead of N/C = (qE) ∆r V Inside the two plates, the E-field is uniform (i.e. No need for integration)

∴ W = qEd Where d is the separation between the plates.

Determination of the electronic charge, e For a stationary droplet : (Millikan’s oil drop experiment 1906-1913) V F = qE mg = qE = q • Oil droplets emitted from a narrow nozzle, due to friction, d negatively acquire negative charges. mgd charged q = • The droplets are kept in a uniform electric field. V mg The mass of the droplets is easily calculated from the oil’s + + + + + + + + + radius (hence its volume) and its density. Fel _ _ _ The results on several hundreds of droplets show that q = Ne. OIL d + _V i.e. a multiple of the electronic charge.

E Fgr EE From his measurements, Millikan concluded that: ______e = 1.6x10-19 C

© Z. Altounian 9 The Capacitor The charge on the capacitor is defined to be the charge on its positively charged plate. A pair of conducting surfaces separated by a gap constitute a capacitor. The amount of charge is directly proportional to the potential difference of the battery. If a capacitor is connected to a battery, after a very short time each plate acquires a charge. Q ~ V

_ + _ +Q + -Q Q = CV _ + _ + _ + _ Where C is the capacitance of the capacitor. + Capacitance refers to the “capacity to accumulate charges” If C is large, so is Q for a given V.

+ – Units for Capacitance: C/V ≡ Farad [F] V

The capacitance C depends on the construction of the capacitor.

• The larger the area of the plates, the more charges can be A put on the plate. i.e. Q ∝ A C = ε0 d d

• The smaller the separation, d, Area = A + _ + _ _ + + between the plates, the greater + _ _ + __ + _ the attraction between the ++ _ Where the proportionality constant, ε0, is the permittivity of oppositely charged plates. + _ vacuum. This will allow more charges -12 2 2 to be deposited on each plate. ε0 = 8.85 x 10 C /Nm i.e. Q ∝ 1/d V

Dielectrics Fill the space between the plates with a dielectric material. +Q -Q What happens if the space between the capacitor plates is filled _ + with a dielectric? _ + _ + _ Q = CEd _ + E 0 _ A dielectric is an insulator (to maintain the charges ±Q on the + Where C is the capacitance with the _ _ _ plates) composed of polarized molecules → + + + _ dielectric material inserted between + _ + _ the two plates. Charge the capacitor by connecting its plates to a battery with + d a potential difference of V. Then, disconnect the battery. The charge on the capacitor plates will not change. +Q -Q The E-field inside the capacitor decreases as the small field of _ + each molecule is opposite in direction to E . i.e. E < E + _ Q = C0V = C0E0d 0 0 + E0 _ Where C0 is the capacitance while in vacuum + _ + _ and E0 is the E-field inside the capacitor. d

© Z. Altounian 10 Since the amount of charge on each plate remains constant, Material κ C0E0d = CEd E C or 0 = ≥1 Vacuum 1 E C0 i.e. The capacitance increases when a dielectric material is Air 1.0005 inserted between the capacitor plates. C Glass 5-9 = κ C0 κ is called the dielectric constant of the material or the relative Paper 3-7 permittivity. The permittivity of the dielectric is ε = κε0. Water 80 In general, A C = ε 0κ d TiO2 100

Example : Construct a 20 pF capacitor using 2 plates with Symbol for the capacitor: dimensions 4 cm x 6 cm. a) What should be the separation between the plates? (+) (-) or Electrolytic capacitor C = κε A κ = 1 d

ε A d = C (8.85 × 10−12 C2 / N ⋅ m2)(0.04 m)(0.06 m) = (20 × 10−12 F)

=1.06×10−3 m

b) If the capacitor is charged to 0.33 µC. What is the E-field Energy stored in a capacitor between the plates? Calculate the work done in moving a charge from one plate to the other.

Q = CV = CEd For the very first charge, q , V = 0, so W = 0. Q 1 1 ⇒ E = As more charges are put on Cd the plates the potential increases and more work (0.33×10−6 C) = must be done to move charges. (20×10−12 F)(1.06×10−3 m) For the very last charge,

WN = qNV 7 = 1.55 ×10 V / m 0 + V V Average potential = = 2 2

© Z. Altounian 11 Other forms for the energy stored in a capacitor If the total charge is Q = q1 + q2 + ... + qN 1 ⎛ V ⎞ 1 PE = QV then the total work done = Q⎜ ⎟ = QV 2 ⎝ 2 ⎠ 2 1 1 which is the potential energy stored in a charged capacitor. = (CV)V = CV 2 2 2 1 1 ⎛ Q ⎞ 1 PE = QV = Q⎜ ⎟ PE = QV 2 2 ⎝ C ⎠ 2 1 Q 2 PE = 2 C

This stored PE can be released by electrically connecting the two plates.

Electric energy density d Example : Calculate the energy stored in a 600 µF capacitor that is charged to a voltage of 400 V. 1 1 ⎛ A ⎞ PE = CV2 = ⎜κε ⎟()Ed 2 2 2 0 d ⎝ ⎠ ← area = A 1 2 1 2 PE = κε 0E Ad Energy = CV 2 2

Ad = volume of the capacitor = 1 (600 × 10−6 F)(400 V)2 2 PE 1 2 Energy density = = κε 0E Ad 2 = 48 J This is a general result. If we know the E-field in a region, we can easily calculate the electric energy density in that region. A car battery ≈ 106 J

Cathode Ray Oscilloscope “CRO” Magnitude of the electric force: Fy = eE = may eE a = Principle of CRO: Deflection of an electron beam by an y m electric field. E is constant between the two charged plates + + + + + + + + + + ⇒ constant acceleration _ F e y From kinematics, the upward deflection of the electron is 0 1 1 eE E x y = v t + a t 2 = t 2 y ∝ E ______oy 2 y 2 m

The electric field E is downwards. (negative y-direction) The E-field is produced by the potential difference, V, applied to the plates. The electric force on the electron is upwards. (y-direction) y ∝ V i.e. Using the CRO, one can measure potential differences.

t t

Screen shows a constant voltage of 3 V Bio-signals