topic 9: motion in fields 9.3 electric field, potential and energy

9.3.1 Define electric potential and electric potential energy.

9.3.2 State and apply the expression for electric potential due to a point charge.

9.3.3 State and apply the formula relating electric field strength to electric potential gradient.

9.3.4 Determine the potential due to one or more point charges.

9.3.5 Describe and sketch the pattern of equipotential surfaces due to one and two point charges.

9.3.6 State the relation between equipotential surfaces and electric field lines.

9.3.4 Solve problems involving electric potential energy and electric potential.

Topic 9: Motion in fields9.3 Electric field, potential and energy

Define electric potential and electric potential energy.

You are probably asking yourself why we are spending so much time on fields.

The reason is simple: Gravitational and electric fields expose the symmetries in the physical world that are so intriguing to scientists.

The Physics Data Booklet shows this symmetry for Topic 6 and Topic 9:


FYIBoth forces are governed by an inverse square law.Mass and charge are the corresponding physical quantities.


We define the electric potential energy at a point P as the amount of work done in moving a charge q from infinity to the point P.

Without going through a rigorous proof, we recall that work = force distance. From Coulomb’s law F = kq1q2/r2

∆EP = Fr

= (kq1q2/r2)r

= kq1q2/r


FYINote the two forms of the proportionality const.

∆EP = kq1q2/r electric potential energy

where k = 8.99×109 N m2 C−2

or ∆EP = (1/40)q1q2/r

0 = 8.85×10-12 C2 N-1

m−2


Since at r = the force is zero, we can dispense with the ∆EP and consider the potential energy EP at each point in space as absolute.


EP = kq1q2/r electric potential energy

where k = 8.99×109 N m2 C−2

or EP = (1/40)q1q2/r

0 = 8.85×10-12 C2 N-1

m−2

EXAMPLE: Find the electric potential energy between two protons located 3.010-10 meters apart.

SOLUTION: Use q1 = q2 = 1.610-19 C. Then

EP = kq1q2/r

= (9.0109)(1.610-19)2/ 3.010-10

= 7.710-19 J.


The potential V (not energy) at any point P in any field is found in the following way:


potential V =

Energy needed to bring test object from infinity to P

Test object interacting with the field

FYIFor the gravitational field the test object is a mass and the potential is measured in J kg-1.For the electric field the test object is a charge and the potential is measured in J C-1.

∆V = ∆EP/m gravitational potential

∆V = ∆EP/q electric potential


From our formula for electric potential energy ∆EP = kQq/r

we see that

∆V = ∆EP/q

= (kQq/r)/q

= kQ/r.


FYIAgain, since at r = the potential energy is zero, we can dispense with the ∆V and consider the potential V at each point in space as absolute.

V = kq/r electric potential where k = 8.99×109 N m2

C−2

Note that there is no (-) in the formula as in V = -Gm/r.

In the case of the electric potential the sign comes from the charge.

State and apply the expression for electric potential due to a point charge.


V = kq/r electric potential where k = 8.99×109 N m2

C−2

PRACTICE: Find the electric potential at a point P located 4.510-10 m from a proton.SOLUTION: q = 1.610-19 C so that V = kq/r = (8.99109)(1.610-19)/(4.510-10) = 3.2 J C-1 (Which is 3.2 V.)PRACTICE: If we place an electron at P what will be the electric potential energy stored in the proton-electron combo?SOLUTION: From ∆V = ∆EP/q we see that ∆EP = q∆V = (-1.610-19)(3.2) = 5.110-19 J (Which is 3.2 eV.)

State and apply the formula relating electric field strength to electric potential gradient.

The electric potential gradient is the change in electric potential per unit distance. Thus the EPG = ∆V/∆r.

Recall the relationship between the gravitational potential gradient and the gravitational field strength g:

Without proof we state that the relationship between the electric potential gradient and the electric field strength is the same:


g = -∆V/∆r gravitational potential gradient

E = -∆V/∆r electric potential gradient

FYIIn the US we speak of the gradient as the slope.In IB we use the term gradient exclusively.

State and apply the formula relating electric field strength to electric potential gradient.


E = -∆V/∆r electric potential gradient

PRACTICE: The electric potential in the vicinity of a charge changes from -3.75 V to -3.63 V in moving from r = 1.80×10-10 m to r = 2.85×10-10 m. What is the electric field strength in that region?

SOLUTION: E = -∆V/∆r E = -(-3.63 - -3.75)/(2.85×10-10 - 1.80×10-10)

E = -0.120/1.05×10-10 = -1.14×109 V m-1 (or N C-1)FYIMaybe it is a bit late for this reminder but be careful not to confuse the E for electric field strength for the E for energy!

Determine the potential due to one or more point charges.

Since electric potential is a scalar, finding the electric potential due to more than one point charge is a simple additive process.


EXAMPLE: Find the electric potential at the center of the circle of protons shown. The radius of the circle is the size of a small nucleus, or 3.010-15 m.

SOLUTION: Because potential is a scalar, it doesn’t matter how the charges are arranged on the circle. Only the distance matters.

For each proton r = 3.010-15 m. Then each charge contributes V = kq/r so that

V = 4(9.0109)(1.610-19)/3.010-15

= 1.9106 N C-1 (or 1.9106 V).

r

Determine the potential due to one or more point charges.


EXAMPLE: Find the change in electric potential energy in moving a proton from infinity to the center of the previous nucleus.

SOLUTION: Use ∆V = ∆EP/q and V = 0:

∆EP = q∆V

= (1.610-19)(1.9106)

= 3.0 10-13 J

Converting to eV we have

∆EP = (3.0 10-13 J)(1 eV / 1.6 10-19 J)

= 1.9106 eV = 1.9 MeV.

r

FYIThis is how much work we would need to do to add the proton to the nucleus.

FYIGenerally equipotential surfaces are drawn so that the ∆Vs for consecutive surfaces are equal.Because V is inversely proportional to r the consecutive rings get farther apart as we get farther from the mass.

Describe and sketch the pattern of equipotential surfaces due to one and two point charges.

Equipotential surfaces are imaginary surfaces at which the potential is the same.

Since the electric potential for a point mass is given by V = kq/r it is clear that the equipotential surfaces are at fixed radii and hence are concentric spheres:


q

equipotential surfaces

Describe and sketch the pattern of equipotential surfaces due to one and two point charges.


EXAMPLE: Sketch the equipotential surfaces due to two point charges.

SOLUTION: Here is the sketch of two negative charges:

Here is the sketch of two positive charges:

qq

FYIThese surfaces are identical to those of two masses. The difference will be in the electric dipole (having no gravitational counterpart).

qq

State the relation between equipotential surfaces and electric field lines.


EXAMPLE: Sketch the equipotential surfaces in the vicinity of a dipole.

SOLUTION: If you recall that equipotential surfaces are perpendicular to the electric field lines the sketches are very simple:

The surfaces are the dashed ellipses.

FYINote that the red E-field is everywhere perpendicular to the blue equipotential surfaces.


In 3D the point charge looks like this:PRACTICE: Explain how the 3D negative point charge picture would differ from that of the positive point charge.

SOLUTION:Simply reverse the E-field arrows so that they point towards the charge instead of away from it.The equipotential surfaces would be the same.


The positive point charge.

State the relation between equipotential surfaces and electric field lines.PRACTICE: Sketch the equipotential surfaces and the E-field surrounding a line of positive charge.

SOLUTION:


State the relation between equipotential surfaces and electric field lines.PRACTICE: Draw a 3D contour map of the equipotential surface surrounding a negative charge.

SOLUTION:


FYIIf you think of the equipotential surface as a hill, a positive test charge placed on the hill will “roll” downhill.

PRACTICE: Identify this equipotential surface.

SOLUTION:

This is a dipole.

The positive charge is the peak and the negative charge is the pit. FYIThe E-field is a conservative field just as the g-field is.





EXAMPLE: Use the 3D view of the equipotential surface to interpret the electric potential gradient E = -∆V/∆r.

SOLUTION: We can choose any direction for our r value, say the y-direction:

Then E = -∆V/∆y.This is just the gradient (slope) of the surface.

Thus E is the (-) gradient of the equipotential surface.

∆y

∆V

Solve problems involving electric potential energy and electric potential.


Equipotentials are perpendicular to E-field lines and they never cross each other!



From E = -∆V/∆r we see that the closer the Vs are, the bigger the E.



From conservation of charge, charge can be neither created nor destroyed.It can, however, be separated into (+) and (-) in equal quantities.Thus, the (-) was separated from the ground, leaving behind an equal amount of (+).

+ + + + + + + + + + + + + + +



+ + + + + + + + + + + + + + +

The electric field strength E at any point in space is equal to the force per unit charge at that point.

If in doubt, find the applicable formula and translate it into words. “E = F/q.”



+ + + + + + + + + + + + + + +

Just recall that between parallel plates the E-field is uniform (equally spaced), and it points from (+) toward (-).



= q/A = 20 C / 7106 m2 = 2.8610-6 C m-2.

0 = 8.85×10-12 C2 N-1

m−2 (from Data Booklet).

Then E = /0 = 2.8610-6 / 8.85×10-12 = 3×105 Vm-1.



is the same everywhere on the cloud.

0 is the same in air as in free space.

The ground and the cloud’s lower surface are flat and parallel.



Use E = -∆V/∆r and ignore the sign:

Then ∆V = E∆r = (3×105)(500) = 1.5108 V!



I = q/t = 20 C / 2010-3 s = 1000 A.

∆EP = q∆V = (20)(1.5108) = 3109 J.

Or…P = IV = (1000)(1.5108) = 1.51011 W. E = Pt = (1.51011)(2010-3) = 3109 J.



The E-field points from more (+) to less (+).Use E = -∆V/∆r and ignore the sign:E = ∆V/∆r = (100 V – 50 V) / 2 cm = 25 V cm-1.



Electric potential at a point P in space is the amount of work done in bringing a charge from infinity to the point P.



The E-field points toward (-) charges.The E-field is ZERO inside a conductor.

Perpendicular to E-field, and spreading.



From E = -∆V/∆r we see that the bigger the separation between consecutive circles, the weaker the E-field.

You can also tell directly from the concentration of the E-field lines.



From V = kq/r we see that V is negative and it drops off as 1/r.

V is biggest (-) when r = a.

V = kq/a

V is ZERO inside a conductor.



V = kq/r

V = (9.0109)(-9.010-9)/(4.5 10-2) = -1800 V.



It will accelerate away from the surface along a straight radial line.Its acceleration will drop off as 1/r2 as it moves away from the sphere.



q = -1.610-19 C.∆EP = q∆V. V0 = -1800 V.

From Vf = kq/r we haveVf = (9.0109)(-9.010-9)/(0.30) = -270 V.

∆EP = q∆V = (-1.610-19)(-270 - -1800) = -2.410-16 J.

∆EK + ∆EP = 0 ∆EK = -∆EP = 2.410-16 J. EKf – EK0 = 2.410-16 J.

0

(1/2)mv2 = 2.410-16.

(1/2)(9.1110-31)v2 = 2.410-16.

v = 2.3107 ms-1.



|E| = ∆V/∆r = (80 – 20) / 0.1 = 600.



V = kQ/r

At any point on the y-axis V = 0 since r is same and paired Qs are OPPOSITE.

On the x-axis V 0 since r is DIFFERENT for the paired Qs.

topic 9: motion in fields 9.3 electric field, potential and energy

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