Question Number 58614 by tanmay last updated on 26/Apr/19
$${with}\:{reference}\:{to}\:{book}\:\:{i}\:{am}\:{posting}\:{some}\:{basic} \\ $$$${question}\:{for}\:{students}. \\ $$$$\left.\mathrm{1}\right){two}\:{charged}\:{coducting}\:{sphere}\:{of}\:{radius} \\ $$$${r}_{\mathrm{1}\:} {and}\:{r}_{\mathrm{2}} \:{are}\:{positvely}\:{charged}\:{separated} \\ $$$${at}\:{a}\:{distance}\:{d}\:{from}\:{each}\:{other}. \\ $$$${force}\:{of}\:{interaction}\:{is}\:{F} \\ $$$${which}\:{is}\:{correct}\:{answer}\: \\ $$$$\left.{a}\right){F}=\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon}\frac{{q}_{\mathrm{1}} {q}_{\mathrm{2}} }{{d}^{\mathrm{2}} } \\ $$$$\left.{b}\right){F}>\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon}\frac{{q}_{\mathrm{1}} {q}_{\mathrm{2}} }{{d}^{\mathrm{2}} } \\ $$$$\left.{c}\right){F}<\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon}\frac{{q}_{\mathrm{1}} {q}_{\mathrm{2}} }{{d}^{\mathrm{2}} } \\ $$$$\left.{d}\right){none}\:{of}\:{these} \\ $$$$\left.\mathrm{2}\right){same}\:{question}\:{as}\:\mathrm{1}\:{but}\:{one}\:{is}\:+{ve}\:{q}_{\mathrm{1}} \:{and} \\ $$$${another}\:{is}\:−{ve}\:{q}_{\mathrm{2}} \:{then}\:{which}\:{option}\:{is}\:{correct} \\ $$$$\left.{a}\right){F}={F}_{{columb}} \\ $$$$\left.{b}\right){F}>{F}_{{columb}} \\ $$$$\left.{c}\right){F}<{F}_{{columb}} \\ $$$$\left.{d}\right){none}\:{of}\:{these} \\ $$$${pls}\:{give}\:{answer}\:{with}\:{explanatiin}.. \\ $$
Answered by Kunal12588 last updated on 27/Apr/19
$${i}\:{think} \\ $$$${ans}\:\mathrm{1}\:−\:\left({b}\right) \\ $$$${ans}\:\mathrm{2}\:−\:\left({b}\right) \\ $$
Commented by tanmay last updated on 27/Apr/19
Answered by sridhar nayak last updated on 09/Jun/19
$${first}\:{question}\:{answer}\:{is}\:\mathrm{1}.{Besause}\:{force}\:{between}\:{two}\:{charged}\:\:{sphere}\:{is}\:{equal}\:\:{to}\:{that}\:{formula}. \\ $$