Physics Courseware Electromagnetism

Transcription

1 Phsics Courseware Electromagnetism Electric potential Problem.- a) Fin the electric potential at points P, P an P prouce b the three charges Q, Q an Q. b) Are there an points where the electric potential is zero? Inicate them in the figure approimatel) an eplain our reasoning. c) Are there an points where the electric fiel is zero? Inicate them in the figure approimatel) an eplain our reasoning. ) What is the total potential energ store b the three charges? Solution: a) The electric potential is a scalar, so we onl nee to a the contribution from each charge: At point P Q Q Q Q Q Q volts At point P Q Q Q Q Q Q ) ) volts

2 At point P Q Q Q Q Q Q volts b) For the electric potential to be zero we have to be closer to the negative charge than to the positive ones. This is because the sum of potentials has to be zero, so: 0 Q Q Q The points that satisf the equation are shown in figure. Figure Figure c) For the electric fiel to be zero the two components of the vectors woul nee to be zero. Assuming an origin at the position where the negative charge is, the electric fiel for a point at,) is given b: ) ) ) / / / ) ), ) ), ), E There are two conitions for the electric fiel to be zero: ) ) ) 0 ) ) / / / E an ) ) ) 0 ) ) / / / E Figure shows where these two points are.

3 ) The total potential energ store b the three charges: Energ Q Q QQ QQ J Problem.- Fin the electric potential at the center of a flat is of raius with a circular hole in the mile of raius r, nowing that the is has a uniform surface charge ensit σ πσ Solution: πσ r) r Problem.- Fin the electric potential at point P, which is on the ais of smmetr of the two uniforml charge rings shown in the figure. Q.5µC, Q µc, 0.5m, 0.48m, X0.m Solution: Since electric potential is a scalar an all the charges in the ring are at the same istance it is the same as if all the charge of the ring where at a point. 6 6 Q Q πε o πε o

4 Problem 4.- A sphere of raius m is locate with its center at the origin of coorinates an has a charge of Q 4nC uniforml istribute over its surface. Another sphere of raius 0.m has its center on the -ais at a istance of D0.4m from the origin of coorinates an has a charge of Q -nc also uniforml istribute over its surface. Calculate the electric potential at point A0,0.5m,0) Solution: There are two contributions, the one ue to the large sphere is Q an the one ue to the smaller sphere the small sphere. Q, where is the istance between point A an the center of 9 9 Q Q So the potential is: volts Q Q Notice that the contribution to the potential ue to the large sphere is, not 0. 5 Problem 5.- A spherical shell has internal raius an eternal raius an contains a uniform istribution of charge with ensit ρ.µc / m Calculate the potential at the center of the shell.

5 Solution: We realize that the electric fiel when r< is zero, so the potential at the 0 center is the same as at a istance so: ) ) Also, outsie the shell the electric potential is the same as if all the charge were locate at the center of the sphere recall Gauss s law) so: Q ) 4 Where Q π )ρ Given this, all we nee to o is fin the ifference in potential between an, which we can o integrating the electric fiel. ) ) Er The electric fiel can be obtaine b integration: 4 E π ρ r / r ) Then we calculate the potential: ) ) π ρ r / r ) r ) 4 4 r πρ r 4 π ) 4 4 ρ πρ πρ ) 0) π ρ 5,000 volts Alternative wa: Divie the shell in thin laers of raius r an thicness r, so the surface area of a laer is 4πr an the volume is 4π r r, then the charge of each laer is 4πr ρr 4 πr ρr an will prouce a potential at the center of 4πρrr, finall we r integrate from to to fin the total: 4πρ rr πρ ) Problem 6.- Fin the electric fiel at point P a istance h.m above the center of a square of sie L.4m an constant charge ensit σ.4µc / m

6 Solution: We learne before that the electric fiel is E σ S. A.) where S.A. is the soli angle that the surface presents to point P. 4π The soli angle in this case is S. A., which ou can calculate b smmetr if ou 6 realize the square an point P are one face an the center of a cube. 4π Then the electric fiel is: E σ 64,00 /m 6 Problem 7.- A sphere of raius cm is locate with the center at the origin of coorinates an has a charge of 4nC uniforml istribute over its surface. Another sphere of raius 6cm has its center on the -ais at a istance of D5cm from the origin of coorinates an has a charge of -nc also uniforml istribute over its surface. Calculate the electric potential at point B5,5,0) Solution: The electric potential at point B5,5,0): In this case we nee to calculate the contribution to the electric potential ue to both spheres. When outsie a spherical istribution of charges we can calculate the potential contribution of the sphere as if all its charge were locate at the center, an so the electric potential at point B is: q q Notice that for electric potential we o not nee to eal with vectors. Potential is just a scalar an the values are either positive or negative epening on the sign of the charges.

7 Problem 8.- Calculate the electric potential at point C ue to the uniforml istribute charge Q on the semicircle of raius. Solution: In problems where there is a continuous istribution of charges the stanar proceure to fin electric fiel or potential is: - Divie the continuous istribution in small pieces ifferentials) that can be treate as point charges. - Calculate the fiel or potential prouce b the ifferential. Be careful that in the case of electric fiel ou will nee to calculate components inepenentl. - All istances, charges an angles shoul be written in terms of the variables chosen. - Integrate over those variables. The problem above falls precisel in the categor that we just mentione. So let s ivie the charge in small pieces. To o this, notice that the charge is uniforml istribute, so the linear ensit of charge is: charge Q λ length π Then, we ivie the arc in ifferentials as shown in the figure: The length of the ifferential of arc is Q Q q λl λ θ θ θ π π l θ an the ifferential of charge is Now we are rea to fin the electric potential. Notice that this ifferential of charge, taen as if it were a point charge, contributes an electric potential equal to:

8 Q θ Qθ q π π Qθ Qπ An we now integrate to fin the potential: π π The result is the same as if all the charge ha been at a point a istance awa. This is not surprising because all the charge is inee at the same istance. π / π / Q