Now suppose that we have a situation where the charge is not distributed uniform

Question

Now suppose that we have a situation where the charge is not distributed uniformly along the bar. The bar is 4000cm long and is charged with 3.600ncC. We want to calculate the electric field strength at point p that is 5.000cm away from the center of the bar. Assume that the charge density varies linearly along the bar, that is lambda (x) = cx Where c is a constant. The total charge on the bar, Q. is distributed nonuniformly along the length of the bar . Of course I don't need to tell you that Q = lambda (x) dx Where L is the length of the bar . You can assume that lambda is constant in each little cell, but it is value in each cell is different. The charge in cell is Delta Q(x) = lambda (x) Delta x = cx Delta x, Where Delta x = L/7 and x is measured from the left end. Please use scientific notation and carry out all calculations to 3 decimal places. A small element of charge , dq = lambda dx and lambda = cx Where c is the slope of the line. The total charge is related to the charge density by Q = lambda (x) dx = cxdx. Use this to calculate c. Shown your work. Fill in the table. Shown a sample of your work.

Explanation / Answer

Q1. Q=integration of c*x*dx with x=0 to x=L

=0.5*c*x^2 , with x=0 to x=L

=0.5*c*(L^2-0^2)=0.5*c*L^2

given that L=0.04 m

Q=3.6*10^(-9) C

then c=4.5*10^(-6) C/m^2

Q2.

as we are assuming the value of lambda is constant in each cell, we can take the value of lambda as the value of labda at the center of the cell.

so center of the 7 cells are at a distance of L/14,3*L/14, 5*L/14,7*L/14,9*L/14,11*L/14,13*L/14

from that we can find lambda in each cell as lambda=c*distance (use c found from question 1)

then charge delta(Q) can be found as integration of lambda*dx , with x varying between boundary of each cell

so for example, for the first cell,

boundary varies from x=0 to x=L/7

lambda=c*L/14

then delta(Q)=integration of (c*L/14)*dx from x=0 to x=L/7

=(c*L/14)*x, from x=0 to x=L/7

=(c*L/14)*(L/7 - 0)=c*L^2/98

similarly we can find for each of the cells.

value of r1 is the distance from the center of the cell to point P.

delta(E) is the magnitude of electric field due to delta(Q) which can be calculated as

9*10^9*delta(Q)/r^2

having defined all the quantities, finally we can produce the table as below:

lambda delta(Q) r1 r1^2 delta(E)

(*10^(-8) (*10^(-11) (*10^(-3) (*10^3)

1.285714 7.346939 0.067143 4.508163 0.146673
3.857143 22.040816 0.061429 3.773469 0.525690
6.428571 36.734694 0.055714 3.104082 1.065089
9.000000 51.428571 0.050000 2.500000 1.851429
11.571429 66.122449 0.044286 1.961224 3.034339
14.142857 80.816327 0.038571 1.487755 4.888889
16.714286 95.510204 0.032857 1.079592 7.962193

note:

the bracketed multiplier is used to make the table more readable.

it can be used as follows:

if first entry in lambda table is 1.285714, and the multiplier is 10^(-8), then actual value of lambda is 1.285714*10^(-8)

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