Answer to Question #282665 in Electricity and Magnetism for danish

Question #282665

1)Two-point charges q1=+9μC and q2=−4μC are separated by a distance L =10 cm, see Fig. below. Find the point at which the resultant electric field is zero.

2)Two identical small spheres of mass m and charge q hang from non-conducting strings, each of length L. At equilibrium, each string makes an angle θ with the vertical, see Fig. below (a) When θ is so small that tanθ ≈ sin θ, show that the separation distance r between the spheres is 𝑟=(𝐿𝑞2/2𝜋𝜖0𝑚𝑔)13. (b) If L =10 cm, m=2 g, and r =1.7 cm, what is the value of q?


1
Expert's answer
2021-12-27T08:02:04-0500

1. The field from each charge is


E1=kq1(rx)2, E2=kq2(x)2.E_1=\dfrac{kq_1}{(r-x)^2},\\\space\\ E_2=\dfrac{kq_2}{(x)^2}.

Somewhere at distance x from the second charge the field is zero:


E1+E2=0, kq1(rx)2+kq2(x)2=0. x=0.2 cm (behind charge 2),x=0.04 cm (between the charges).E_1+E_2=0,\\\space\\ \dfrac{kq_1}{(r-x)^2}+\dfrac{kq_2}{(x)^2}=0.\\\space\\ x=-0.2\text{ cm (behind charge 2)},\\ x=0.04\text{ cm (between the charges)}.

2. Three forces act on each sphere: the force of gravity, tension, and electrical force:


Tcos(θ/2)=mgT=mgcos(θ/2), Tsin(θ/2)=kq2r2r2=kq2Tsin(θ/2)=kq2mgtan(θ/2), kq2mg[tan(θ/2)]=kq2mg[sin(θ/2)]=r2.T\cos(\theta/2)=mg→T=\dfrac{mg}{\cos(\theta/2)},\\\space\\ T\sin(\theta/2)=\dfrac{kq^2}{r^2}→r^2=\dfrac{kq^2}{T\sin(\theta/2)}=\dfrac{kq^2}{mg\tan(\theta/2)},\\\space\\ \dfrac{kq^2}{mg[\tan(\theta/2)]}=\dfrac{kq^2}{mg[\sin(\theta/2)]}=r^2.

On the other hand:


sin(θ/2)=r2L, 2Lkq2mgr=r2, r=(Lq22πϵ0mg)1/3.\sin(\theta/2)=\dfrac r{2L},\\\space\\ \dfrac{2Lkq^2}{mgr}=r^2,\\\space\\ r=\bigg(\dfrac{Lq^2}{2\pi\epsilon_0mg}\bigg)^{1/3}.


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