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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
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: displacement
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The work depends on only the initial position and final position of q o, and has nothing to do with the path. q0q0
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---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,
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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:
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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 ,
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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
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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.
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4. Electric potential Z Definition E-potential difference: --Describe the character of E-field. Work:
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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
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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
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( 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
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dq a ( 3 ) The E-potential of charged body Divide q many of dq For any dq : For entire charged body : charge element
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Integrating for charged body Caution !! This method can be used for the finite distribution charged body.
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[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
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(1)(1) (2)(2)
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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
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Use the superposition principle of E-potential-- superposition method. Integrating for charged body superposition method
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[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
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Method -- superposition method q dq
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Discussion (1) (2) If the charged body is a half circle, ? 0 R
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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 :
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R + + + ++ + + + q (1) For any point P outside the sphere ( ) P r (2) For any point inside the sphere surface
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+ + + + + q + + + + + + + R P The sphere is equipotential. 分段积分分段积分
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Conclusion ( ) ( )
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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
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superposition method : Integrating for charged body It’s very complex !!
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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.
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[Example 4] Calculate the E- potential distribution of an infinite line with uniform charge(the linear density isλ). Solution Use definition method
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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
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§8-6 Equipotential surface and Potential gradient 1. Equipotential surface --the potential has the same value at all points on the surface.
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Positive point charge Electric dipole Dash line-- equipotential surface Real line-- -line
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A parallel plate capacitor +++++++++
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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
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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
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aa bb Prove : Assume there are two equipotential surface U a, U b then -line points on the direction of the increase of the potential.
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UaUa UbUb UcUc r2r2 r1r1 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 :
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2. Potential gradient Equipotential surface - line If the distribution of U is known, How can we calculate ?
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(1) Special example : uniform field unit positive charge a b, the work done by E-force: C aa b Express that the ’s direction is the direction of U decrease. q=1
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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.
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(2) Any E-field: definition : potential gradient grad U The normal direction of equipotential surface, point on U increasing = U= U U U
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In Cartesian coordinate system
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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
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[Example 1] Calculate the E-field of an electric dipole. P( )
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[example 2] calculate the E-field on the axis of the round charged plate using its potential gradient.
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Solution The potential of any ring on the axis:
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i.e., the potential on the axis:
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§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 qq +q+q l --electric dipole moment (1) E-dipole in the uniform field
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The E-dipole will rotate under the action of the torque until the direction of is the same with F F qq +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
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(2) E-dipole in non-uniform field F1F1 F2F2 qq +q+q 11 l 22 --Move to the area of larger E-field --rotating
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2. A moving charged particle in a uniform E-field q E x=U
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q x y
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