Collaboration (lst name only De: Friday Septeeslbet 22 st 9am In clase oe Assign
ID: 3279613 • Letter: C
Question
Collaboration (lst name only De: Friday Septeeslbet 22 st 9am In clase oe Assignments Masilber in Allen Hall I. (0 pts) Logistics (a) (0 pt) Print-out this sheet and attach it to your write-up as the first page. (b) (0 pe) Attach your write-up of the online hw at the end of the written hw; required for credit. (e) (0 pt) Staple your homework together. Some people folded the corners and it didn't stay together 2, (2S pts) Line of charge on the p-axis at P (0, b). We'll repeat part of what I did lecture to set up problem 3 (a) (5 pts) Set-up. Draw a line of length L (draw it as a long thin rectangle) along the y-axis centered at (0,0). It should look vertical. We will consider the line chopped up in to N pieces. Furthermore, we will approrimate each little piece to be a point charge at the middle of the piece. For N 3, daw the chopped up boundaries and the point charges. The location of each piece is (01), where i 0.1, , N-i labels the piece. So Yi means the value on the y-axis for the in polnt. Write the width of the piece dy in terms of the symbolic parameters given so far. Write the expression for the chunk of charge da. It should contain some or all of the following terms: , dy, aud i. (b) (5 pts) Differential element. Consider a point P on the z-axis with coordinate (a, b). Write the expression for dE and dE. Express cos and sin in terms of the parameters such as z and a. (c) (10 pts) Analytic solution & limit. For a special point on the y-axis P-(0, b) where b >· integrate the above expression to find the analytic form of E. (d) (5 pts) ForbL, showy that the equivalent system to produce the same É is a point charge at the origin with total charge Q- 3. (25 pts) Line of charge at an arbitrary point P-(a,b). (a) (10 pts) Let's find the E field at P using the point-charge approximation using the followingExplanation / Answer
2. a. so for length L
divided into 3 parts
location of centres of three segmets are (0,L), (0,0), (0,-L)
where dy = L/3
charge on each chunk dq = lambda*dyi = lambda*dy
b. for the point on x axis , coordinates = (x,0)
so, dEix = k*xdq/(x^2 + yi^2)^3/2
dEiy = k*yi*dq/(x^2 + yi^2)^3/2
c. for a point on y axis, (0,b), b > L/2
dEx = 0
dEy = k*dq/(b - y)^2 = k*l;ambda*dy/(b - y)^2
integrating
Ey = k*lambda[1/(b - L/2) - 1/(b + L/2)] = k*lambda[4L/(4b^2 - L^2)]
d. for b > > L
Ey = k*lambda[1/(b - L/2) - 1/(b + L/2)] = k*lambda[4L/(4b^2 - L^2)]
4b^2 > > L^2
Ey = k*lambda[L/(b^2)]
Ey = kQ/b^2
where Q = lambda*L