Question

A very small sphere with positive charge 5.00uC is released from rest at a point 1.20cm from a very long line of uniform linear charge density 3.00 uC/m.. . What is the kinetic energy of the sphere when it is 4.50cm from the line of charge if the only force on it is the force exerted by the line of charge?

Answer 1

Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval (−L,0] for some large number L. The voltage V as a function of x on the interval (0,∞) is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element dx′, we have λ=dqdx′.V(x)=1/4πϵ0∫linedq/r=λ/4πϵ0∫−L0dx/x−x′=λ/4πϵ0(ln|x+L|−ln|x|)

Answer 2

Based on the calculations, the kinetic energy of this very small sphere is equal to 0.36 Joules.

Given the following data:

• Charge, q = 5.00 uC = 5 × 10⁻⁶ C.

• Initial point = 1.20 cm to m = 0.012 m.

• Final point = 4.50 cm to m = 0.045 m.

• Charge density, λ = 3.00 uC/m = 3 × 10⁻⁶ C/m.

How to calculate the kinetic energy?

Since the only force acting over the very small sphere is the force exerted by the line of charge, the work done is given by:

W = ΔK = qV    ......equation 1.

Also, an electric field of infinite line charge is given by:

$V = \frac{\lambda}{2\pi \epsilon_o } ln\frac{r_2}{r_1}$      .....equation 2.

Substituting eqn. 2 into eqn. 1, we have:

$K.E=qV\\\\K.E =q(\frac{\lambda}{2\pi \epsilon_o } ln\frac{r_2}{r_1})\\\\K.E=5 \times 10^{-6} \times ( \frac{3 \times 10^{-6}}{2 \times 3.142 \times 8.85 \times 10^{-12} })\times ln(\frac{0.045}{0.012})\\\\K.E=5 \times 10^{-6} \times ( \frac{3 \times 10^{-6}}{5.56 \times 10^{-11}})\times ln(3.75)\\\\K.E = 5 \times 10^{-6} \times ( 53956.83) \times 1.32$

K.E = 0.36 Joules.

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