Electric potential §8-5 Electric potential Electrostatic field does work for moving charge --E-field possesses energy 1.Work done by electrostatic force  Let test charge q 0 moves a  b along arbitrary path in the E-field set up by point charge q. The work done by electrostatic force=? q0q0

: displacement 

The work depends on only the initial position and final position of q o, and has nothing to do with the path. q0q0

---has nothing to do with path  When q o moves in the E-field set up by charges’ system q 1, q 2,  q n,

 When q o moves in the E-field set up by charged body, Conclusion: -- the work has nothing to do with path 2. Circular theorem of electrostatic field When q 0 moves along a closed path L ， Electrostatic force is conservative force. E-force does work:

3. Electric potential energy -- The E-potential energy when q 0 at point a and b. 、 Electrostatic field is conservative field. The work done by electrostatic force = the decrease of the electric potential energy Circular theorem of electrostatic field q 0 moves in E-field a  b ，

Notes (1) E P a is relative quantity. If we want to decide the magnitude of E P when q 0 at a point, we must choose zero reference point of E-potential energy. Z The work done by E-force for q 0 =- increment of E- potential energy. Z--zero E-potential energy point Z

The choice of zero E-potential energy point ：  Choose zero point at  when the charge distribution is finite.  Choose zero point at the finite distance point when the charge distribution is infinite. (2) E P is scalar. It can be positive, negative or zero. (3) E P depends on E-field and ， it belongs the system.

4. Electric potential Z Definition E-potential difference: --Describe the character of E-field. Work:

5.Calculation of E-potential (1) The E-potential of a point charge q Take U  =0 ， then the E-potential of a point a : The distance from q to a

Discussion U r + U r  If q>0 ， U>0 for any point in the space. when r  ， U  U(  )  0  If q<0 ， U<0 for any point in the space. when r  ， U  U(  )  0

（ 2 ） The E-potential of a system The system of point charges ： q 1,q 2, ,q n qiqi a riri q1q1 r1r1  --superposition principle of E-potential

dq a （ 3 ） The E-potential of charged body Divide q  many of dq For any dq ： For entire charged body ： charge element

Integrating for charged body Caution ！！ This method can be used for the finite distribution charged body.

[Example 1] Four point charges q 1 ＝ q 2 ＝ q 3 ＝ q 4 ＝ q is put on the vertexes of a square with edge of a respectively. Calculate (1)The E-potential at point 0. (2)If test charge q 0 is moved from  to 0, how much work does the E-force do? 6. Examples of calculating E-potential

（1）（1） （2）（2）

Definition method Z Integrating for path Calculate the E-potential set up by a charged body Two methods  Use the definition of E-potential as the distribution of is known. – definition method

 Use the superposition principle of E-potential-- superposition method. Integrating for charged body superposition method

[Example 2] Calculate the E- potential on the axis of a uniform charged ring. q 、 R are known. Solution Method  Use the E-field distribution of the ring that was calculated before. The direction: along x axis

Method  -- superposition method q dq

Discussion (1) (2)  If the charged body is a half circle, ？ 0 R

R q [Example 3] Calculate the E-potential distribution of a charged spherical surface E-field distribution: （  ） R r （  ） Solution  Definition method Zero potential point ： 

R q (1) For any point P outside the sphere （  ） P r  (2) For any point inside the sphere surface

q R P The sphere is equipotential.  分段积分分段积分 

Conclusion （  ） （  ）

The distribution curve of E-potential The distribution curve of E-field ： E  R R r r O O 8 8 r2r2 r 1 1

 superposition method ： Integrating for charged body It’s very complex ！！

Conclusion  When the E-field distribution is symmetry and it can be calculated by using Gauss’s Law conveniently, it is simpler to calculate potential by using  definition method.  When the E-field distribution is not symmetry and it can’t be calculated by using Gauss’s Law conveniently, it is simpler to calculate potential by using  superposition method.

[Example 4] Calculate the E- potential distribution of an infinite line with uniform charge(the linear density isλ). Solution Use  definition method

How to choose the zero potential point? Choose any point b as zero potential point  When r p 0.  When r p >r b ， U <0. Finite distance to the charged line

§8-6 Equipotential surface and Potential gradient 1. Equipotential surface --the potential has the same value at all points on the surface.

Positive point charge Electric dipole Dash line-- equipotential surface Real line-- -line

A parallel plate capacitor

Prove ： Assume q 0 moves along equipotential surface a  b ， then ： The properties of equipotential surfaces  No net work is done by the E-field as a charge moves between any two points on the same equipotential surface. a b q0q0

Prove ： Assume on an equipotential surface, the field at point P is then ： q 0 moves along equipotential surface, q0q0 P  -lines are always normal to equipotential surfaces

aa bb Prove ： Assume there are two equipotential surface U a, U b then  -line points on the direction of the increase of the potential.

UaUa UbUb UcUc r2r2 r1r1  the density of equipotential surfaces shows the magnitude of E-field. Prove ： Assume there is a family of equipotential surfaces U a 、 U b 、 U c 、  E1E1 E2E2 then ：

2. Potential gradient Equipotential surface - line If the distribution of U is known, How can we calculate ?

(1) Special example ： uniform field unit positive charge a  b, the work done by E-force: C  aa  b Express that the ’s direction is the direction of U decrease. q=1

q=1 moves a  c, the work done by E-force ： ’s component at the direction of ’s component at any direction = the negative magnitude of the rate of U change with distance on that direction.

(2) Any E-field: definition ： potential gradient grad U The normal direction of equipotential surface, point on U increasing = U= U  U U

In Cartesian coordinate system

a.the magnitude of at any point = the maximum of the rate of U change with distance on that point. the direction of is perpendicular to the equipotential surface at that point and along the the direction of U decreasing. b. In the space that U is constant,  U = 0 ，  E = 0 c. E is not sure = 0 in the space of U= 0. U is not sure = 0 in the space of E= 0. conclusion

[Example 1] Calculate the E-field of an electric dipole. P( )



[example 2] calculate the E-field on the axis of the round charged plate using its potential gradient.

Solution The potential of any ring on the axis:

i.e., the potential on the axis:

§8-7 E-Force Exerted on a Charge 1. The E-force exerted on a charged particle The torque exerted on the dipole: F F qq +q+q  l --electric dipole moment (1) E-dipole in the uniform field

The E-dipole will rotate under the action of the torque until the direction of is the same with F F qq +q+q  l Assume the potential energy of E-dipole is zero at the position  =90 0 E p (  =90 0 ) =0 Then the potential energy of E-dipole at any orientation is

(2) E-dipole in non-uniform field F1F1 F2F2 qq +q+q 11 l 22 --Move to the area of larger E-field --rotating

2. A moving charged particle in a uniform E-field q E  x=U

q x y