ch 25.3 – potential and pot. energy of point charge pretend a point charge q is sitting out in the...

Ch 25.3 – Potential and Pot. Energy of Point ChargeCh 25.3 – Potential and Pot. Energy of Point Charge

• Pretend a point charge q is sitting out in the universe.

• It generates an E-field, for which we can write a potential function.

• Let’s find the electric potential at a point r away from the charge.

B

A

AB sdEVVV


• We know the E-field of the point charge:

• So…

sdrr

qksdE e

ˆ

2

rr

qkE e ˆ

2


• Also, the magnitude of r-hat is always equal to 1, so…

sdrr

qksdE e

ˆ

2

coscosˆˆ dssdrsdr

cos2

dsr

qksdE e


• But, dscosθ is the projection of ds onto r, which is just the differential change in r.

drds cos

drr

qksdE e 2


• In other words, if our little steps ds cause a change in r, then E dot ds will take some value.

B

A

r

r

e

B

A

AB r

drqksdEVV

2

drr

qksdE e 2


B

A

r

r

e

B

A

AB r

drqksdEVV

2

B

A

r

reAB r

qkVV


ABeAB rrqkVV

11

The electric potential difference between two points, A and B, due to a point charge.

Notice, the potential difference is path independent.

It only depends upon the radial-distance-change from A to B.


ABeAB rrqkVV

11The electric potential a distance r away from a point charge, setting zero potential an infinite distance from the charge.

It’s our job to decide where the electric potential is zero.

It makes the math easiest if we zero the potential at rA = infinity relative to the point charge.

r

qkV e

0

0


The electric potential a distance r away from a point charge, setting zero potential an infinite distance from the charge.

r

qkV e

A plot showing the electric potential some distance from a positive point charge.

We set the potential at 0 when r = infinity, so the electric potential is small far away and grows as you move toward the charge.

Keep in mind, this actually happens in all 3 dimensions, but we can only represent 2 here.

Reminds you of a hill, right?


• To get the electric potential from a group of i point charges, we simply sum their individual potentials (principle of superposition).

i i

ie r

qkV


The electric potential surrounding a symmetrical dipole.

r

q

r

qkV e

For instance…

The electric potential due to a symmetrical dipole.


• Recall:

• We’re putting VA out at infinity now.

• So, if we move a charge q2 in from infinity toward another point charge q1, the electric potential energy will change by an amount:VqU

12

21

r

qqkU e


• In other words, if we started two charges, q1 and q2, infinitely far apart, and then we brought them together to a final spacing of r12, we’d need to supply at least this much energy:

• Of course, that’s assuming the charges don’t accelerate.

• U represents the electric potential energy of the two-charge system.

12

21

r

qqkU e


• If we have more than two charges, we need to account for all interactions in the system.

• For instance, the electric potential energy of a three-charge system would be:

• This is the minimum work you’d need to do to start the charges infinitely far apart and move them together to this final state.

23

32

13

31

12

21

r

qq

r

qq

r

qqkU e

Charge q1 = 2.00 μC and is at the origin.

Charge q2= -6.00 μC and is at ordered pair (0, 3.00)m.

(a) Find the total electric potential due to these charges at the point P, located at (4.00, 0)m.

(b) Find the change in PE of the system when a third charge, q3 = 3.00 μC, moves in from infinity to point P.

EG 25.3 The electric potential due to 2 point chargesEG 25.3 The electric potential due to 2 point charges

Remember this equation?

It says a finite change in potential will occur if you move from A to B and an E-field exists in that region.

Stands to reason that we should be able to get the E-field in the region if we know the change in electric potential between A and B.

Ch 25.4 – Getting the E-field from the E-potentialCh 25.4 – Getting the E-field from the E-potential

B

A

sdEV

If we only take one little step, ds, then the change in potential is infinitesimally small:


sdEdV

If we only take one little step, ds, then the change in potential is infinitesimally small:

Let’s make it simple. Pretend the E-field only points in the x direction, and therefore only has one component Ex. Then:


sdEdV

dxEsdE x

So, for this 1-D example,


dxEdV x

dx

dVEx

This is actually valid in all three dimensions:

In other words, if we know the spatial-rate-of-change of the V function in one of the coordinate directions, then we know the component of the E-field in that dimension.


x

VEx

y

VEy

z

VEz

Experimentally, electric potential and position can be measured easily (using a voltmeter and a meter stick). So, you can determine the E-field at some point by measuring the electric potential at several positions in the field and making a graph of the results.


x

VEx

y

VEy

z

VEz


x

VEx

y

VEy

z

VEz

E

E

E

E

E

E

A dipole consists of two charges, equal in magnitude and opposite sign.

They’re separated by a distance 2a as shown.

The dipole lies along the x axis, centered at the origin.

(a) Calculate the electric potential at point P on the y axis.

(b) Calculate the electric potential at point R on the +x axis.

(c) Calculate V and Ex at a point way down the x axis.

EG 25.4 The electric potential of a DipoleEG 25.4 The electric potential of a Dipole

ch 25.3 – potential and pot. energy of point charge pretend a point charge q is sitting out in the...

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