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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. - PowerPoint PPT Presentation
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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