Chapter 17 Electrical Energy and Current

17 – 1 Electric Potential

• Electric potential energy (EPE): the potential energy associated with an object’s position in an electric field

• Electrical potential energy is a component of mechanical energy.

ME = KE + PEgrav + PEelastic + PEelectric

Electrical potential energy can be associated with a charge in a uniform field.

ΔPEelectric = qEd magnitude of change in PE

q = chargeE = electric fieldd = displacement from a reference

point

Review

• In the case of gravity, the gravitational field does work on a mass

• WAB=F d = mgd

• ΔPE = W AB= mgΔh

• In the case of a charge moving in an electric field

• As the positive charge moves from A to B, work is done

• WAB = F d = q E d

• ΔPE = W AB= q E Δd

Energy and Charge Movements

Direction of movement

+ charge - charge

Along E Loses PE Gains PE

Opposite E Gains PE Loses PE

Electric potential energy for a pair of point charges

r

qqkPE c

21

Note: The reference point for zero potential energy is usually at

Example

• What is the potential energy of the charge configuration shown?

q 3 = 1.0 c

q 2 = - 2.0 cq 1 = 2.0 c

0.30

m

0.30 m

0.30 m

Given:

q 1 = 2.0 x 10-6 C

q 2 = - 2.0 x 10-6 C

q 3 = 1.0 x 10-6 C

r = 0.30 m (same for all)

k= 9.0 x 109 n•m2/C2

The total potential energy is the algebraic sum of the mutual potential energies of all pairs of charges.

• PET = PE12 + PE23 + PE 13

• = k q1q2 + kq2q3 + kq1q3

r r r

• = k [(q1q2) + (q2q3 ) + (q1q3)]

r

= [(9.0 x 109 n•m2/C2)/ 0.30] x [(2.0 x 10-6 C)(- 2.0 x 10-6 C) + (- 2.0 x 10-6 C)(1.0 x 10-6 C) +(2.0 x 10-6 C)(1.0 x 10-6 C)]

=- 0.12 J

Electric Potential

• Electric potential: the electric potential energy associated with a charged particle divided by the charge of the particle

• symbol for electric potential = V

V = PE/q

• SI unit = volt (V) • 1Volt = 1 Joule/Coulomb

Potential Difference

• Potential Difference equals the work that must be performed against electric forces to move a charge between the two points in question, divided by the charge.

• Potential difference is a change in electric potential.

chargeelectric

energypotentialelectricinchangedifferencepotential

q

PEV electric

Chapter 17Potential Difference, continued

• The potential difference in a uniform field varies with the displacement from a reference point.

∆V = –Ed Potential difference in a uniform electric field

E = electric fieldd = displacement in the field

Sample Problem

• A proton moves from rest in an electric field of 8.0104 V/m along the +x axis for 50 cm. Find

• a) the change in in the electric potential,

• b) the change in the electrical potential energy, and

• c) the speed after it has moved 50 cm.

• a) V = -Ed = -(8.0104 V/m)(0.50 m) = -4.0104 V

• C) Ei = Ef

• KEi+PEi = KEf + PEf,

since KEi=0

KEf = PEi – PEf = -PEPE

1/2 m1/2 mppvv22 = - = -PEPE

v = v = 2 2 PE/mPE/m = 2(6.4x10= 2(6.4x10-15-15 J)/1.67x10 J)/1.67x10-27-27 kg)=2.8x10 kg)=2.8x1066 m/s m/s

b) PE = q V = (1.610-19 C)(-4.0 104 V) = -6.4 10-15 J

Chapter 17Potential Difference, point charges

• At right, the electric poten-tial at point A depends on the charge at point B and the distance r.

• An electric potential exists at some point in an electric field regardless of whether there is a charge at that point.

Chapter 17

• The reference point for potential difference near a point charge is often at infinity.

• Potential Difference Between a Point at Infinity and a Point Near a Point Charge

r

qkV c

Electric Potential of Multiple Point Charges- Superposition

• The total electric potential at point “a” is the algebraic sum of the electric potentials due to the individual charges.

Problem Solving with Electric Potential (Point Charges)

• Remember that potential is a scalar quantity– So no components to worry about

• Use the superposition principle when you have multiple charges– Take the algebraic sum

• Keep track of sign– The potential is positive if the charge is positive and

negative if the charge is negative

• Use the basic equation V = kcq/r

Example: Finding the Electric Potential at Point P

5.0 C = = -2.0 C

V1060.3)m0.4()m0.3(

)C100.2()C/Nm1099.8(

,V1012.1m0.4

C100.5)C/Nm1099.8(

3

22

6229

2

46

2291

V

V

Superposition: Vp=V1+V2

Vp=1.12104 V+(-3.60103 V)

=7.6103 V