Question

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.

Image for The figure below shows a finite line charge with linear charge density of ? and total length L. The point P sh

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
P = _____ N/C î + _____ N/C j
0 0
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Answer #1

dt dE

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)

image from custom entry tool

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


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Answer #2

AM 9 Carn Co 속 쓰cesg

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Answer #3

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?

Here the charge 'q' is placed in axial position, hence the electric field at the position of charge due to line charge is
   E   =   (1/4pe0) * 2 * ? * l / (d2 - l2)
   here linear charge density      ?   =   total charge on line charge (Q) / length (L)
                                                   =   5.49 mC / 0.531
                                                   =   1.034 * 10-2   C/m
   2 * l   =   length of thin rod
            =   L
   =>   l   =   L / 2
            =   0.531 / 2
            =   0.2655   m
   =>      E   =   9.0 * 109 * 2 * 1.034 * 10-2 * 0.2655 / (0.4012 - 0.26552)
                  =   5.472 * 108   N/C
   Force of charge q         F   =   q * E
                                          =   1.6 * 10-19 * 5.472 * 108

                                          =   8.755 * 10-11    N
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Answer #4

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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?

Here the charge 'q' is placed in axial position, hence the electric field at the position of charge due to line charge is
   E   =   (1/4pe0) * 2 * ? * l / (d2 - l2)
   here linear charge density      ?   =   total charge on line charge (Q) / length (L)
                                                   =   5.49 mC / 0.531
                                                   =   1.034 * 10-2   C/m
   2 * l   =   length of thin rod
            =   L
   =>   l   =   L / 2
            =   0.531 / 2
            =   0.2655   m
   =>      E   =   9.0 * 109 * 2 * 1.034 * 10-2 * 0.2655 / (0.4012 - 0.26552)
                  =   5.472 * 108   N/C
   Force of charge q         F   =   q * E
                                          =   1.6 * 10-19 * 5.472 * 108

                                          =   8.755 * 10-11    N
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