1Phy 213: General Physics IIIChapter 24: Electric PotentialLecture NotesElectric Potential Energy1. When charge is in an electric field, the electric force exerted upon it, as it moves from one point (A) to another point (B) can do work:2. The total work performed to move a charge from point A to B in an electric field is then:3. Since the electric force is a conservative force, the work performed is equal to the difference in electric potential energy between points A & B:EdW=F ds = qE ds⋅ ⋅  B BAB EA AW = dW = F ds = q E ds⋅ ⋅∫ ∫ ∫  AB A BW = U - U2Electric Potential1. The electric potential energy for a charge at a point in space can be expressed in per unit charge, this is the electric potential at this location:a. In this manner, V associated with a position can be expressed without concern for the specific charge presentb. The SI units are joules/coulomb (J/C), called the volt (V)2. The work performed on an electric charge is related to the change in electric potential (∆V) and its charge (q):or3. Expressing W in terms of V we can combine them to obtain a relation between E and V:orEUV = qAB-W =q V∆ABWV = -q∆VE = -s∂∂BABAW- = V = - E dsq∆ ⋅∫Potential due to a Point ChargeConsider a point charge, q1. The electric field is radial from q:2. The electric potential a distance r from q is:This relationship is extremely useful for evaluating more complex charge geometries…r r2o oq dr qV=- E ds = - = 4 r 4 rπε πε∞ ∞⋅∫ ∫ˆr2oqE = i4 rπεrqV3Potential due to a Charged SphereFor a hollow conducting sphere with radius R & charge, q:1. The electric field outside R is :2. The electric potential for points outside R is :3. The electric field inside the sphere is:4. The electric potential inside the sphere is constant:rqGaussian sphereRr r2o oq dr qV=- E ds = - = 4 r 4 rπε πε∞ ∞⋅∫ ∫2oqE = 4 rπεE = 0f i f iVE = - = 0 V -V= E ds= 0 V = V= constant s∂⇒ ⋅ ⇒∂∫Potential due to an Electric Dipole1. Consider an electric dipole with charge separation d and dipole moment p2. The electric field for a dipole is:3. The potential decreases inversely with r2:dipole3pE = 2 roπε   r3r3 2pV = - E ds = dr2 rp cos dr p cosV = - = 2 r 4 roo oπεθ θπε πε∞∞ ⋅ ⋅   ⋅ ⋅  ∫ ∫∫4Potential due to a Line of Charge1. For an line of charge, of length L and charge density λ:2. The electric field is described by:3. The electric potential is:ˆr22λ LE = i4Ly + y4oπε         ( )L1/202 2o o2 2o1 dq dxV = dV = = 4 r 4x + yL + L + yV = ln4 yλπε πελπε       ∫ ∫ ∫+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + xVλ0LrydqPotential due to a Charged Disc1. The electric potential due to a thin ring of charge, dq at point P is:2. The electric potential at point P is:()( ) ( )1 12 22 2 2 2o oo2 r drdq rdrdV = = =4 d 24 y +r y +rσ πσπε επε   ( )( )1212R2 22 2o o0rdrV = dV = = y +R - y2 2y +rσ σε ε            ∫ ∫PxRrdVφφφφ5Potential between 2 Parallel Charged PlatesTwo oppositely charged conducting plates, with charge density σ:1. The electric field between the plates is constant:2. The corresponding electric potential is linear between the plates:oVE = = - xσε∂∂+q-qxxo o0V=- E dx = dx = xσ σε ε⋅∫