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22 Pages

### Chapter 24b

Course: PHY 2048 PHY, Spring 2009
School: University of Florida
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Word Count: 1854

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24: Chapter Electric Potential Part B ! ! ! ! ! ! ! Electric potential energy and potential. Calculating the potential from the field. Potential due to a point charge. Potential due to a group of point charges. Potential due to an electric dipole. HITT question. Potential due to a continuous charge distribution Line of charge. Charged disk. Ch. 24: Electric Potential - Part B 1 September September 16, 2009...

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24: Chapter Electric Potential Part B ! ! ! ! ! ! ! Electric potential energy and potential. Calculating the potential from the field. Potential due to a point charge. Potential due to a group of point charges. Potential due to an electric dipole. HITT question. Potential due to a continuous charge distribution Line of charge. Charged disk. Ch. 24: Electric Potential - Part B 1 September September 16, 2009 Electric Potential Energy initial state q1 q2 final state q1 q2 q3 q3 Ui Uf If a system of charges changes its configuration from an initial state i to a different final state f, the electrostatic force does work W on the particles. The work done is path independent (conservative force). U = U f Ui = W September September 16, 2009 Ch. 24: Electric Potential - Part B 2 Electric Potential ! The electric potential V is the potential energy per unit charge at a point in an electric field. U V= q ! ! The electric potential is a scalar (not a vector). Electric potential difference between two points. U V = Vf Vi = q !! !! U = W = F d = qE d !! V = E d September September 16, 2009 Ch. 24: Electric Potential - Part B 3 Calculating the Potential from the Field ! We can calculate the potential difference between any two points i and f in an electric field if we know the electric field vector E all along any path connecting those points. ! ! First we calculate the work done on a positive test charge +q0 by an arbitrary (nonuniform) electric field E as the charge moves along the path from i to f. At any point on the path, an electrostatic force q0E acts on the charge as it moves through a differential displacement ds. !! ! ! dW = F ds for F = q0 E September September 16, 2009 !! dW = q0 E ds Ch. 24: Electric Potential - Part B 4 Calculating the Potential from the Field ! We find the total work by summing (integrating) the differential works as the charge moves through all the displacements ds along the path. !! W = dW = q0 E ds i f W V = Vf Vi = q0 f !! Vf Vi = E ds The potential difference between any two points i and f in an electric field is the negative of the!line!integral (integral along a particular path) of E ds from i to f. September 16, 2009 Ch. 24: Electric Potential - Part B 5 i Calculating the Potential from the Field ! If we set Vi=0, then !! V = E ds ! f The above equation also gives the potential at any point f relative to the zero potential at infinity. i September 16, 2009 Ch. 24: Electric Potential - Part B 6 Calculating the Potential from the Field Charge q0 is moved between points i and f in an uniform electric field E. Points i and f lie on the same electric field line. Calculate the potential difference Vf-Vi for the path icf. Verify that you will get the same result for the direct path if! !! ! ! Vc Vi = E ds = 0 since E ds , and =90 c if f !! Vf Vc = E ds = E cos( 90 ) ds = E (sin ) dcf d = E (sin ) = Ed sin September 16, 2009 c c Ch. 24: Electric Potential - Part B 7 Potential due to a Point Charge ! ! ! !! V = E ds i f ! The positive point charge q produces an electric field E and an electric potential V at point P (at distance R from the point charge). We move a test charge q0 (shown at a distance r from the point charge) from point P to infinity, to find the potential. The electric field E is directed radially outward from the fixed charge q. For the integration we take the direct path, a line that extends radially from the fixed charge through point P to infinity. Ch. 24: Electric Potential - Part B 8 September 16, 2009 Potential due to a Point Charge ! ! The differential displacement ds of the test charge along its path has the same ! direction as E . !! E ds = E cos ds = E cos 0 ds = Eds !! Vf Vi = E ds = Edr i R ds = dr since the path is radial f Vf = V ( ) = 0 and Vi = V ( R ) = V 0 V = Edr where E = R September 16, 2009 1 q 4 0 r 2 for the electric field of a point charge 9 Ch. 24: Electric Potential - Part B Potential due to a Point Charge 0 V = 1 q 4 0 r 2 R q dr = q 4 0 r R 1 2 dr q 1 = r = 4 4 0 R 0 q 1 0 = 4 0 R R V= 1 q 4 0 r general result, r is the radial distance from the point charge, q can be any point charge (+,-,0) A positively charged particle produces a positive electric potential, while a negatively charged particle produces a negative electric potential. September 16, 2009 Ch. 24: Electric Potential - Part B 10 Potential due to a Point Charge A computer-generated model of V(r) for points on the (x,y) plane. The (positive) charge is at (0,0). V (r) = 1 q 4 0 r The same equation is valid for the electric potential outside or on the external surface of a symmetric spherically charge distribution (using the shell theorem for the electric field). V= 1 q 4 0 r q is the net charge of the sphere or shell, and r is the radial distance from the center of the charge distribution Ch. 24: Electric Potential - Part B 11 September 16, 2009 Potential due to a Group of Point Charges ! The net electric potential V for a group of point charges is found by applying the superposition principle. V = Vi = 1 ! n 1 4 0 r 1 n qi i ! This is an algebraic sum, the signs of the charges are important. Much easier to calculate than a vector sum, such as when calculating the net electric field. September 16, 2009 Ch. 24: Electric Potential - Part B 12 Potential due to a Group of Point Charges ! Examples. The charges are two protons. Rank the potentials for the three arrangements. q1 q2 q 1 1 V = V1 + V2 = + = + r2 4 0 d D 4 0 r1 1 The electric potential is a scalar and does not depend on the orientation. The three potentials are equal! September 16, 2009 Ch. 24: Electric Potential - Part B 13 Potential due to a Group of Point Charges ! 2Q r The charges are located at the corners of a square of side d=0.24 m. Calculate the potential at point P at the center of the square. 4 4 4Q 1 4 qi 1 V = Vi = 1 d P q 1 4 4 0 1 r = 4 0 q r 1 i i = q1 + q2 + q3 + q4 = 2Q + 4Q +Q Q = 6Q -Q +Q r = d / 2 and if Q=1.5 nC, then 6 Q 2 ( 8.99 109 Nm2 /C2 )( 6 1.5 109 C ) 2 V= = = 477 V 4 0 d 0.24 m September 16, 2009 Ch. 24: Electric Potential - Part B 14 Potential due to an Electric Dipole ! ! ! An electric dipole with charges +q and q at a distance d apart. Calculate the potential at an arbitrary point P (at a distance r from the center of the dipole). Let r(+) and r(-) are the distances from P to the +q and q charges, respectively; and V(+) and V(-) are the potentials produced by the +q and q charges at point P, respectively. V = V( + ) + V( ) q q = + 4 0 r( + ) r( ) 1 q r( ) r( + ) = 4 r r 0 () (+) 15 September 16, 2009 Ch. 24: Electric Potential - Part B Potential due to an Electric Dipole ! For distances r much further than the size d of the dipole, we can use the following approximations. r( ) r( + ) d cos r( ) r( + ) r 2 ! p V= V= q 4 0 1 4 0 r( ) r( + ) r( ) r( + ) p cos r2 q 4 0 d cos r2 where p=qd is the magnitude of the electric dipole moment. September 16, 2009 Ch. 24: Electric Potential - Part B 16 HITT Question How many hours per week do you spend studying for PHY2049? A) 0-2 hours; B) 2-4 hours; C) 4-6 hours; D) 6-8 hours; E) >8 hours. September September 16, 2009 Ch. 24: Electric Potential - Part B 17 Electric Potential ! Potential due to a continuous charge distribution. ! dq For continuous charge distribution, we can not use summation to find the electric potential, but must use integration. ! We choose a differential element of charge dq, determine the potential dV at point P due to dq (treated as a point charge), and integrate over the entire distribution. dV = 1 dq 4 0 r then V = dV = 1 4 0 dq r 18 September 16, 2009 Ch. 24: Electric Potential - Part B Electric Potential due to a Line of Charge ! dq L A thin nonconducting rod of length L has a positive charge of uniform linear density . Calculate the electric potential V due to the rod at point P, a perpendicular distance from the left end of the rod. dq = dx where = q / L r = d2 + x2 dV = 1 dq 4 0 r = 1 dx 4 0 x 2 + d 2 ( ) 1/2 September 16, 2009 Ch. 24: Electric Potential - Part B 19 Electric Potential due to a Line of Charge V = dV = 0 L 1 4 0 dx x2 + d2 dx V= 4 0 dq 0 L L V= 4 0 V= 4 0 l n L + x 2 + d 2 0 ( l n L + ( x2 + d2 L2 + d 2 ) ln d ) L 20 V>0 since q>0. V= 4 0 L + L2 + d 2 l n d September 16, 2009 Ch. 24: Electric Potential - Part B Electric Potential of a Charged Disk ! ! ! A plastic disk of radius R, charged on its top surface to a uniform surface charge density . Find the potential V at point P on the central axis of the disk at a distance z from the center. We choose a differential element consisting of a flat ring or radius R and radial width dR. Its charge has a magnitude dq = dA = d R ' 2 = ( 2 R ' )( dR ' ) ! ( ) All points of the charged element are the same distance r from point P. r = z 2 + R '2 September 16, 2009 Ch. 24: Electric Potential - Part B 21 Electric Potential of a Charged Disk dV = 1 dq 4 0 r = 1 4 0 ( 2 R ' )( dR ' ) z 2 + R '2 V = dV = 2 0 0 R R ' ( dR ' ) z 2 + R '2 z>0 R 2 V= z + R '2 0 2 0 dy y =2 y V= 2 0 ( z 2 + R2 z ) y = R '2 + z 2 September 16, 2009 Ch. 24: Electric Potential - Part B 22
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