Question

Figure 22-47 a shows a nonconducting rod with a uniformly distributed charge +Q. The rod forms a 10/22 of circle with radius R and produces an electric field of magnitude Earc at its center of curvature P. If the arc is collapsed to a point at distance R from P (Figure 22-47b), by what factor is the magnitude of the electric field at P multiplied?

Answer 1

Linear charge density of the rod

$$ \lambda=\frac{Q}{L}=\frac{Q}{2 \pi R\left(\frac{10}{22}\right)}=\frac{Q}{0.91 \pi R} $$

Linear charge density in differential form

$$ \begin{array}{l} \lambda=\frac{d Q}{d x}=\frac{Q}{0.91 \pi R} \\ d Q=\frac{Q}{0.91 \pi R} d x \end{array} $$

Now the electric field

$$ \begin{aligned} d E &=k \frac{d Q}{R^{2}}=k \frac{1}{R^{2}} \frac{Q}{0.91 \pi R} d x \\ &=k \frac{Q}{0.91 \pi R^{3}} d x \end{aligned} $$

Now, since length is changing along a circle, we can write an equation that models the change in length in terms of the change in angle and the constant radius of the rod

$$ d x=R d \theta $$

Now electric field

$$ d E=k \frac{Q}{0.91 \pi R^{3}} R d \theta $$

Resolve into components

$$ \begin{array}{l} d E_{x}=\frac{k Q}{0.91 \pi R^{2}} \cos \theta d \theta \\ d E_{y}=\frac{k Q}{0.91 \pi R^{2}} \sin \theta d \theta \end{array} $$

By insgrating above equations

$$ \begin{aligned} E_{x} &=\int_{x / 2}^{-x^{2}} \frac{k Q}{0.91 \pi R^{2}} \cos \theta d \theta \\ &=\frac{k Q}{0.91 \pi R^{2}} \int_{x / 2}^{-\sqrt 2} \cos \theta d \theta \\ &=\frac{k Q}{0.91 \pi R^{2}}(-2) \end{aligned} $$

Similarly y-component

$$ E_{y}=\int_{z / 2}^{-x_{2}} \frac{k Q}{0.91 \pi R^{2}} \sin \theta d \theta $$

$$ =0 $$

Then magnitude of electric field

$$ \begin{aligned} E &=\sqrt{E_{x}^{2}+E_{y}^{2}} \\ &=\sqrt{\left(\frac{k Q}{0.91 \pi R^{2}}(-2)\right)^{2}+0^{2}} \\ &=(2.2) \frac{k Q}{\pi R^{2}} \end{aligned} $$

Now, we must find the magnitude of the electric field after the rod is shrunk down to a point charge. Then the magnitude of an electric field for the point charge with charge \(Q\) and distance \(R\) from point \(P\) is

$$ E=\frac{k Q}{R^{2}} $$

So, the factor by which the magnitude of the electric field shrunk,

$$ =\frac{\frac{k Q}{R^{2}}}{(2.2) \frac{k Q}{\pi R^{2}}}=\frac{\pi}{2.2}=1.43 $$

Answer 2

This is an interesting problem. To start analyze the symmetry of the sphere and realize that there is no net y component of the field. Next, I wouldcalculate λ by diving Q over the circumference, aka 2*π*R/2=R*π. Then I would write the equation for the electric field ∫K*dQ/r^2. However, because wehaveλ, we cand rewrite dQ asλdl. And because this is a semi circle, I would write λdl as λ*2*π*dθ. And, as mentioned earlier, we only worryabout the x component so Ex=∫K*Q*2*π*cosθ*R*dθ/(R*π*r^2) from -π/2 toπ/2, where cosθ is the angle between the x axis and dθ, also note that r^2is a constant of R^2 (yay circles). So Ex=2*(2*K*Q*2/R^2)∫cosθdo from 0 toπ/2 (symmetry). You should be able to solve from there.

Answer 3

1.43