In the figure below, a rod of length L = 8.00 cm has charge q = -4.28 fC which i
ID: 581315 • Letter: I
Question
In the figure below, a rod of length L = 8.00 cm has charge q = -4.28 fC which is uniformly distributed along its length. The labelled point P is a distance a from the end of the rod.
(a) What is the linear charge density of the rod?
What is the the electric field at point P: (use for parts b-c)
(b) Nearby:
i.) for a = 12.0 cm:
ii.) for the same a, but now with the entire rod compressed into a point charge (located at the rod's original midpoint):
(c) Far away:
Repeat the previous part with a = 50 meters.
part (a) ansewer: -5.35e-14
Explanation / Answer
a) linear charge density, lamda = Q/L
= -4.28*10^-15/0.08
= -5.35*10^-14 C/m
b)
i) on the axis of the rod,
E = k*lamda*(1/a - 1/(L+a) )
= 9*10^9*5.35*10^-14*(1/0.12 + 1/(0.08+0.12) )
= 6.42*10^-3 N/c (towards -x axis)
ii) E = k*q/d^2
= 9*10^9*4.2*10^-15/(0.04 + 0.12)^2
= 1.48*10^-3 N/c (towards -x axis)
c) i) on the axis of the rod,
E = k*lamda*(1/a - 1/(L+a) )
= 9*10^9*5.35*10^-14*(1/0.12 + 1/(0.08+0.5) )
= 4.84*10^-3 N/c (towards -x axis)
ii) E = k*q/d^2
= 9*10^9*4.2*10^-15/(0.04 + 0.5)^2
= 1.3*10^-4 N/c (towards -x axis)