Electric Potential

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# Electric Potential - PowerPoint PPT Presentation

Electric Potential. Physics 102 Professor Lee Carkner Lecture 11. PAL #10. q 2 = +4 m C. Find E 1 and E 2 E 1 = 8.99X10 9 (2X10 -6 )/(3 2 ) = 1124 N /C r 2 2 = 3 2 + 4 2 r 2 = 5 m E 2 = 8.99X10 9 (4X10 -6 )/(5 2 ) = 1438 N /C Find q 1 and q 2 q 1 = 0 (right on x-axis)

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## Electric Potential

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Electric Potential

Physics 102

Professor Lee Carkner

Lecture 11

PAL #10

q2 = +4 mC

• Find E1 and E2
• E1 = 8.99X109(2X10-6)/(32) = 1124N /C
• r22 = 32 + 42
• r2 = 5 m
• E2 = 8.99X109(4X10-6)/(52) = 1438N /C
• Find q1 and q2
• q1 = 0 (right on x-axis)
• Can get q2 from triangle
• tan q2 = ¾, q2 = 37 degrees
• q2 is below X axis so is -37

r2 =5 m

3 m

q2

E1

q2

4 m

E2

q1 = +2 mC

PAL #10

q2 = +4 mC

• Find X and Y components
• E1x = E1 cos q1 = 1124 cos 0 = 1124 N/C
• E1y = E1 sin q1 = 1124 sin 0 = 0
• E2x = E2 cos q2 = 1438 cos -37 = 1148 N/C
• E2y = E2 sin q2 = 1438 cos -37 = -865 N/C
• Find total E vector and angle
• Etotal,x = E1x + E2x = 1124 + 1148 = 2272 N/C
• Etotal,y = E1y + E2y = 0 – 865 = -865 N/C
• E2total = E2total,x + E2total,y
• E = (22722 +8652)½
• E = 2431 N/C
• tan qtotal = Etotal,x / Etotal,y = -865 / 2272
• qtotal = -20 degrees
• 20 degrees below + X axis

r2 =5 m

3 m

E1

q2

q2

4 m

E2

q1 = +2 mC

Electrical Potential Energy
• The value of the potential tells us how much potential energy is at that point per unit charge
• V = PE/q or PE = Vq
• 1V = 1 J/C
• Which could be converted into work or kinetic energy
Point Charges and Potential
• Consider a point charge Q, what is the potential for the area around it?
• At infinity the potential is zero
• It can be shown that:

V = ke Q / r

• Q is charge providing potential, q is charge experiencing potential
Finding Potential
• If the potential is provided by an arrangement of charges:
• Total V = V1 + V2 + V3 …
• DV = -Ed
• We don’t know the potential at anyone point, but we can find the difference between two points separated by a distance d
Potential and Energy
• The change in energy is:

DPE = Vfq – Viq = q(Vf-Vi) = qDV

DPE = -W

• Energy could be converted into kinetic energy:

DPE = DKE

KEi + PEi = KEf + PEf

• Particle could speed up or slow down
Problem Solving
• Particle moving between two points of different potential
• Find DV and then DPE (=qDV)
• Add or subtract DPE to initial energy of particle
Signs
• As a positive charge moves with the electric field, the particle gains kinetic energy and loses potential energy
• Like dropping a ball in a gravitational field
• the electric field does positive work
• If a positive charge is forced backwards against an electric field, the particle loses kinetic energy and gains potential energy
• like rolling a ball up a hill
• the field “does” negative work
Potential and Charges
• A negative particle:
• Opposite of positive particle
• A negative particle has the most potential energy at low potential
• The potential and the potential energy are two different things
• Potential at a point is the same no mater what kind of test charge is put there

High potential

Low potential

Work
• Since energy must be conserved:
• DPE + W = 0 or DPE = -W
• Work done by the field is positive if it decreases the potential energy
• The “natural” movement
• When the charge is forced in the “unnatural” direction
• The negative work done by the system is the positive work done on the system
Down

gain KE

lose PE

field does +work

“natural”

Up

lose KE

gain PE

you do work

field “does” negative work

“forced”

For negative particle, everything is backwards

But high and low potential are still in the same place

High potential

+

E

Low potential

Equipotentials

Blue = field = E

Dashed = potential = V

• Each line represents one value of V
• Particles moving along an equipotential do not gain or lose energy
• Equipotentials cannot cross
Next Time
• Homework, Ch 17: P 10, 20, 35, 46

sign of DPE

sign of DV

sign of W

naturally?

+ charge moves with E field

+ charge moves against E field

-charge moves with E field

-charge moves against E field

Electric Potential Chart
The above electric field,
• increases to the right
• increases to the left
• increases up
• increases down
• is uniform
Is it possible to have a zero electric field on a line connecting two positive charges?
• Yes, at one point on the line
• Yes, along the entire line
• No, the electric field must always be greater than zero
• No, but it would be possible for two negative charges
• No, the electric field is only zero at large distances