In the figure below, a nonconducting rod of length L = 8.60 cm has charge ?q = ?
ID: 1587943 • Letter: I
Question
In the figure below, a nonconducting rod of length L = 8.60 cm has charge ?q = ?4.32 fC uniformly distributed along its length.
(a) What is the linear charge density of the rod? C/m
(b) What is the magnitude of the electric field at point P, a distance a = 12.0 cm from the end of the rod? N/C
(c) What is its direction? ° (counterclockwise from the positive x axis)
(d) What is the electric field magnitude produced at distance a = 50 m by the rod? N/C
(e) Repeat part (d) for a particle of charge ?q = ?4.32 fC that replaces the rod. N/C
-9 -a 7Explanation / Answer
(a) Linear charge density of the rod () = q/L = -4.32*10-15/8.6*10-2 = -5.02*10-14 C/m
(b) Electric field in this case is given by E = k*q/a*(a+L) = 9*109*4.32*10-15/0.12*(0.12+0.086) = 1.573*10-3 N/C
(c) Direction would be 1800 counter-clockwise from the +ve x-axis (towards the rod)
(d) Here again the electric field would be E = k*q/a*(a+L) = 9*109*4.32*10-15/50*(50+0.086) = 1.552*10-8 N/C
(e) In this case electric field would be E = k*q/a2 = 9*109*4.32*10-15/50*50 = 1.555*10-8 N/C