In the figure below, a rod of length L = 7.00 cm has charge q = -4.76 fC which i

Question

In the figure below, a rod of length L = 7.00 cm has charge q = -4.76 fC which is uniformly distributed along its length. The labelled point P is a distance a from the end of the rod. What is the linear charge density of the rod? What is the the electric field at point P: (use for parts b-c) for the same a, but now with the entire rod compressed into a point charge (located at the rod's original midpoint): (How do your answers for i-ii compare?) Repeat the previous partwith a = 50 meters. (How do your answers for i-ii compare now?)

Explanation / Answer

a)

here

the charge per unit length is

lambda = q/L

lambda = -4.76 * 10^-15 / 0.07 = -6.8 * 10^-14 C/m

first we write dq for a little length of charge

dq = lambda dx

we then write the r , r vector , r unitvector for the charge dq

r = sqrt( a - x)^2

r vector = (a - x) i

r unitvector = r vector / r = i

we can then compute the x and y components of the electric field

E = integrate(dq / (4 * pie * e0 *r^2) r = integrate from 0 to -L ( lambda dx / (4 * pie * e0(a - x)^2 ) i

E = (lambda i /( 4 * pie * e0) ) * integrate from 0 to -L( dx / (a -x)^2 )

E = ( lambda i / (4 * pie * e0) ) * ( 1 / a - 1 / (a+L))

E = (lambda * L / ( 4 * pie * e0)) * ( 1 / (a*(a + L) ))i

E = ((6.8 * 10^-14 * 0.07) / (4 * 3.14 * 8.854 * 10^-12) ) * ( 1 / (0.12*(0.12 + 0.07)))

E = 0.00187 N/C

the field point to the left

then by using the same formula for 50 m also

E =  ((6.8 * 10^-14 * 0.07) / (4 * 3.14 * 8.854 * 10^-12) ) * ( 1 / (50*(50 + 0.07)))

E = 1.71 * 10^-8 N/C

the rod looks quite pointlike from 50m

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