Two-similar-ball-of-mass-m-attached-by-silk-thread-of-length-a-and-carry-similar-charge-q-assume-is-small-enough-that-tan-sin-to-this-approximation-show-that-X-qa-2pi-0-mg-1-3-w

FacebookTweetPin

Question Number 50994 by peter frank last updated on 23/Dec/18

$$Twosimilarballofmass$$$$mattachedbysilkthread$$$$oflengthaandcarry$$$$similarchargeq.assume\theta is$$$$smallenoughthat$$$$tan\theta \approx sin\theta tothis$$$$approximation,show$$$$thatX={\left(\frac{qa}{2\pi {\epsilon }_{0}mg}\right)}^{\frac{1}{3}}$$$$whereXisdistanceof$$$$separation.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18

$$T=tensioninthread$$$$Tcos\theta =mg$$$$Tsin\theta =F$$$$F=repulsiveforce=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{x^2}$$$$tan\theta =\frac{F}{mg}$$$$sin\theta =\frac{F}{mg}$$$$\frac{x}{2a}=\frac{1}{4\pi {ϵ}_0}\times \frac{q^2}{x^2}\times \frac{1}{mg}$$$$x^3=\frac{{aq}^2}{2\pi {ϵ}_0}\times \frac{1}{mg}$$$$x={\left(\frac{{aq}^2}{2\pi {ϵ}_{0}mg}\right)}^{\frac{1}{3}}$$

Answered by peter frank last updated on 23/Dec/18

$$F=\frac{q^2}{4\pi {\epsilon }_0{r}^2}[q_1=q_2]$$$$\sin \theta =\frac{x}{2a}$$$$\tan \theta =\frac{F}{mg}$$$$F=mg\tan \theta$$$$\tan \theta \approx \sin \theta =\theta$$$$F=mg\frac{x}{2a}\dots \dots \left(i\right)$$$$F=\frac{q^2}{4\pi {\epsilon }_0{r}^2}[r=x]\dots ..\left(ii\right)$$$$mg\frac{x}{2a}=\frac{q^2}{4\pi {\epsilon }_0{x}^2}$$$$x={\left[\frac{q^{2}a}{4\pi {\epsilon }_{0}mg}\right]}^{\frac{1}{3}}$$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

FacebookTweetPin