CourseNotes.5

# CourseNotes.5 - 3.5 Problems Problem 22.25 In ﬁgure 2 two...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3.5 Problems Problem 22.25 In ﬁgure 2, two curved plastic rods, one of charge +q and the other of charge −q , form a circle of radius R = 8.50 cm in an xy plane. The x axis passes through both of the connecting points, and the charge is distributed uniformly on both rods. If q = 15.0 pC , what are the a) magnitude and b) dircction (relative to the positive direction of the x axis) of the electric ﬁeld E produced at P , the center of the circle? Figure 2: Problem 22.25 Solution Part a) As with all of these problems, we begin with the diﬀerential form of the electric ﬁeld. dE = 1 dq 4π 0 r 2 We can relate the diﬀerential element of charge dq to the diﬀerential element of length along the hoop via the linear charge density. The charge density will be the same for the top and the bottom q portion of the ring with opposite signs λ = ± πR . We can use this to describe the diﬀerential element of charge in terms of the diﬀerential element of length dθ. dq = λ R dθ = ± q dθ π We must be careful here. Our goal is to sum up all of the contributions from all of the diﬀerential elements of charge to the ﬁeld at the center of the ring. The electric ﬁeld is a vector quantity however, and so we can not simply sum it up as if it was a scaler. The symmetry of the problem allows us to greatly simplify the problem though. If we consider an element of charge dq located at a position θ (where θ is chosen to be 0 at the positive x axis) then it will have a corresponding piece of charge at π − θ which will cancel out the x portion of the ﬁeld. Hence, only the y component of the ﬁeld will contribute for each diﬀerential piece of charge. The y component of the ﬁeld is a scaler quantity and hence can be integrated in the normal way. dEy = 1 dq q sin θ = ± 2 sin θ dθ 4π 0 R2 4π 0 R2 Integrating gives the total electric ﬁeld q 2π π sin θ dθ − sin θ dθ 4π 2 0 R 2 0 π q π 2π = − cos θ|0 − − cos θ|π 4π 2 0 R 2 q =2 = 2.02 N C π 0 R2 E= 5 ...
View Full Document

## This note was uploaded on 12/05/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online