The figure below shows a charged semi-circular arc. Find the electric field at p
ID: 581131 • Letter: T
Question
The figure below shows a charged semi-circular arc. Find the electric field at point P (the center of curvature) as a function of lambda, R, and theta_1 and theta_2. Then use your result to evaluate E vector_p for each of the following values. Look for symmetries to avoid extra work! You may use R = 4 m for parts, and lambda = 4.10^-6 C/m for parts a-c. theta_1 = theta_2 = pie/6: E vector_p = N/C i^ + N/C j^ theta_1 = theta_2: E vector _p = N/C i^ + N/C j^ (Do your expect this to be larger or smaller than your answer for the previous part Why In the limit that this arc becomes a full ring: E vector _p = N/C i^ + N/C j^ (Make sure that your answer makes intuitive sense!) Repeat part a, except with lambda = -4.10^-6 C/m. E vector_p = N/C i^ + N/C j^ Repeat part b, but now suppose lambda = +4.10^-6 C/m on the upper half (+y) of the arc and -4.10^-6 C/m on the lower half (-y). Repeat part b, but with changing theta_2 ONLY to pie/4: E vector _p = N/C i^ + N/C j^Explanation / Answer
a) E= [k lambda ]/R [ sin theta1 +sin theta 2]
= [k lambda/R ] [ sin pi/6 +sin pi/6]
=[9*109 *4*10-6 /4] [0.5+0.5] =9000 N/C
electric field is zero in y direction because of the symmetry
so Ep = 9000 N/C i + 0 N/C j
b) E= [k lambda ]/R [ sin theta1 +sin theta 2]
= [k lambda/R ] [ sin pi/2 +sin pi/2]
=[9*109 *4*10-6 /4] [1+1] =18000 N/C E= [k lambda ]/R [ sin theta1 +sin theta 2]
electric field is zero in y direction because of the symmetry
so Ep = 18000 N/C i + 0 N/C j
yes, we expected higher value because net charge increased so electric field should be more in x direction
c) E= [k lambda/R ] [ sin theta1 +sin theta 2]
if it becomes full ring, theta1 =theta 2 =pi
E= [k lambda/R ] [ sin pi +sin pi] =[k lambda/R ] *0 = 0
Due to symmetry all the fields will cancel out each other.
Ep = 0 N/C i + 0 N/C j
d) It will be just negative of that in part a
Ep = -9000 N/C i + 0 N/C j
e) Just add another +lambda (quarter circle) on the lower half and symmetrically opposite to the circle (top-right quarter to center) so that this addition has no net effect.
Now we have got a semicircle of charge +4 *10-6 C/m....opened in -y direction
electric field is zero in X direction because of the symmetry
so Ep = 0 N/C i + -18000 N/C j
f) Ex = [k lambda/R ] [ sin pi/2 +sin pi/4]
= 9000 *[ 1+ 0.707] N/C i =15364 N/C i
Ey = - [k lambda/R ] [ sin 0 +sin 3pi/4]
= - 9000*0.707 = - 6363 N/C j
Hence E = 15364 N/C i - 6363 N/C j
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