The figure below shows a finite line charge with linear charge density of λ and total length L. The point P shown is a distance s away from its end.
Please calculate a formula for the electric field at point P, in terms of λ, L and s.
Then use the following values to find it numerically.
λ = +7 μC/m, L = 4 m, s = 3 m
Homework Answers
Consider a small charge element, dt at K. Where KO = t
Angle OPK = θ
The electric field due to this element is:
dE = (1/4πε) [ λdt / (S2 + t2 ) ] cosθ i - (1/4πε) [ λdt / (S2 + t2 ) ] sinθ j
Note : the component along Y-axis is negative, since y-axis is upward.
cosθ = S / (S2 + t2)1/2 , sinθ = t / (S2 + t2)1/2
So, t =
S tanθ => dt = S sec2θ
dθ
dE = (λ/4πε) [S sec2θ / (S secθ)2] cosθ dθ i - (λ/4πε) [ Ssec2θ / (Ssecθ)2] sinθ dθ j
= (λ/4πε) (1/S) cosθ dθ i - (λ/4πε) (1/S) sinθ dθ j
Where i and j are unit vectors along X and Y axis respectively.
Integrating: θ varies from 0 to tan-1 (L/S)
Note: tan-1(L/S) = sin-1 [ L / (L2 + S2)1/2 ] = cos-1 [ S / (L2 + S2)1/2 ]
( Just construct a triangle and verify)
in this image θ = tan-1(L/S)
E = (λ/4πε) (1/S) [ L / (L2 + S2)1/2 ] i - (λ/4πε) (1/S) [ - S / (L2 + S2)1/2 + 1 ] j
= (λ/4πε) (1/S) [ L / (L2 + S2)1/2 ] i - (λ/4πε) (1/S) [ 1 - S / (L2 + S2)1/2 ] j
λ = 7 μ C/m, L = 4 m, S = 3 m
E = 7 x 10-6 x 9 x 109 x (1/3) x [ 4 / 5] i - 7 x 10-6 x 9 x 109 x (1/3) x [ 1 - 3/5] j
= 16.8 x 103i - 8.4 x 103j N/C
Ex = 16.8 K N/C
Ey = - 8.4 K N/C
The figure shows a uniformly charged thin rod of
length L = 0.531 m that has total charge Q = 5.49 mC. What is the
magnitude of the electrostatic force acting on an electron
positioned on the axis of the rod at a distance d = 0.401 m from
the midpoint of the rod?
= 8.755 * 10-11 N
The figure shows a uniformly charged thin rod of length L = 0.531 m that has total charge Q = 5.49 mC. What is the magnitude of the electrostatic force acting on an electron positioned on the axis of the rod at a distance d = 0.401 m from the midpoint of the rod?
= 8.755 * 10-11 N
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