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bloth pin 0 2mont ma 0.18 S For the arrangement de- scribed in Problem 26, calcu- Figure P20.26 Problems 26 late the electric potential at and 27. point B, which lies on the pendicular bisector of the rod a distance babove the x axis. per-

bloth pin 0 2mont ma 0.18 S For the arrangement de- scribed in Problem 26, calcu- Figure P20.26 Problems 26 late the electric potential at and 27. point B, which lies on the pendicular bisector of the rod a distance babove the x axis. per-

College Physics
College Physics
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Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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I seem to be having issues with this app today. I’m gonna try this question again... I need some help working out number 27 in the attached photo. I also included the work I have done so far in another photo. I am trying to find the electric potential at point B. Any help would be appreciated! If possible, any short explanations you can offer would surely help! Thank you!
onstoegs bns Isitnalo ohbela o Problems 691 difference between the point at the center of the ring and a point on its axis a distance 2Rfrom the center? 26 SA rod of length L (Fig. P20.26) lies along the x axis with its left end at the origin. It has a nonuniform charge density A = ax, where a is a positive constant. (a) What are the units of a? (b) Calcu- late the electric potential at A. B es d A 27. S For the arrangement de- scribed in Problem 26, calcu- Figure P20.26 Problems 26 late the electric potential at point B, which lies on the pendicular bisector of the rod a distance babove the x axis. and 27. per- 28. S WA wire having a uniform linear charge density A is bent into the shape shown in Figure P20.28. Find the elec- tric potential at point 0. R 2R 2R Figure P20.28 29. WA uniformly charged insulating rod of length 14.0 cm is bent into the shape of a semicircle as shown in Figure P20.29. The rod has a total charge of –7.50 µC. Find the electric potential at 0, the center of the semicircle.
Transcribed Image Text:onstoegs bns Isitnalo ohbela o Problems 691 difference between the point at the center of the ring and a point on its axis a distance 2Rfrom the center? 26 SA rod of length L (Fig. P20.26) lies along the x axis with its left end at the origin. It has a nonuniform charge density A = ax, where a is a positive constant. (a) What are the units of a? (b) Calcu- late the electric potential at A. B es d A 27. S For the arrangement de- scribed in Problem 26, calcu- Figure P20.26 Problems 26 late the electric potential at point B, which lies on the pendicular bisector of the rod a distance babove the x axis. and 27. per- 28. S WA wire having a uniform linear charge density A is bent into the shape shown in Figure P20.28. Find the elec- tric potential at point 0. R 2R 2R Figure P20.28 29. WA uniformly charged insulating rod of length 14.0 cm is bent into the shape of a semicircle as shown in Figure P20.29. The rod has a total charge of –7.50 µC. Find the electric potential at 0, the center of the semicircle.
meriments - nurselves WI "n on a sphere an the electric fier Problems 691 7.00 x 10-C difference between the point at the center of the ring and a naN m/C? 8.00 x 10-9 C point on its axis a distance 2R from the center? 0 070 0 m CHAPTER 20 (27) 0.030 0 m 26. SA rod of length L (Fig. nvie FOR the arranyement described, calculate the electric potential at poin t B, which lies on the perpendicular bisector of the rod a'distance b above the Xaxis. * divide into infinitesimal *nonuni form elemints of charge da *dv= Ke dar 1. Linear charge density: -ax da= Adx - axdx Y A *S over the entire charye distnibution Q= -*) = x Kea xdx
Transcribed Image Text:meriments - nurselves WI "n on a sphere an the electric fier Problems 691 7.00 x 10-C difference between the point at the center of the ring and a naN m/C? 8.00 x 10-9 C point on its axis a distance 2R from the center? 0 070 0 m CHAPTER 20 (27) 0.030 0 m 26. SA rod of length L (Fig. nvie FOR the arranyement described, calculate the electric potential at poin t B, which lies on the perpendicular bisector of the rod a'distance b above the Xaxis. * divide into infinitesimal *nonuni form elemints of charge da *dv= Ke dar 1. Linear charge density: -ax da= Adx - axdx Y A *S over the entire charye distnibution Q= -*) = x Kea xdx
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