A charged nonconducting rod, with a length of 2.35 m and a cross-sectional area

Question

A charged nonconducting rod, with a length of 2.35 m and a cross-sectional area of 5.88 cm2, lies along the positive side of an x axis with one end at the origin. The volume charge density ? is charge per unit volume in coulombs per cubic meter. How many excess electrons are on the rod if ? is (a) uniform, with a value of -3.66

Explanation / Answer

The volume charge density "D" is defined as the amt of charge "dq" divided by the volume element "dV" it occupies; D = dq/dV dq = DdV To find the amt. of charge in a finite volume "V" integrate; q = INTEGRAL[DdV] In your case the cross-sectional area is comstant so you can write dV = Adx q = (A)INTEGRAL[Ddx] (a) D is constant so the integral is simply; q = (A)(D)INTEGRAL[dx] from x=0 to x = 2.35 m = (5.88x10^-4)*(-3.66x10^-6)*(2.35-0) = -5.057 x10^-9 C charge of electron = -1.6x10^-19 => no. of excess electrons = 5.057x10^-9 / 1.6x10^-19 = 3.16x10^10 (b) D = bx^2 and the integral is; q = (b)(A)INTEGRAL[x^2dx] = (-1.73x10^-6)*(5.88x10^-4)*(x^3)/3 --------(put the limits as x=0 to x=2.35) = (-1.73x10^-6)*(5.88x10^-4)*(2.35^3)/3 = 4.4005 x10^-9 C => no of electrons = 4.4004x10^-9/1.6x10^-19 = 2.7x10^10

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