Question Number 46292 by Umar last updated on 23/Oct/18
$${A}\:+{ve}\:{point}\:{charge}\:{of}\:{magnitude}\:{q}\:{is}\:{located}\:{on}\:{the}\:{y}−{axis}\: \\ $$$${at}\:{a}\:{point}\:{of}\:{y}=+{d}\:{and}\:{another}\:−{ve}\:\:{charge}\:{of}\:{same}\:{magnitude} \\ $$$${is}\:{located}\:{on}\:{the}\:{point}\:{y}=−{d}.\: \\ $$$${A}\:{third}\:+{ve}\:{charge}\:{of}\:{same}\:{magnitude}\:{is}\:{located}\: \\ $$$${at}\:{some}\:{point}\:{on}\:{x}−{axis}. \\ $$$$\:\:\:\left(\mathrm{1}\right)−\:{what}\:{is}\:{the}\:{magnitude}\:{and}\:{direction}\:{of}\:{the}\:{force}\:{on} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{third}\:{charge}\:? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({a}\right)−\:{when}\:{its}\:{located}\:{on}\:{the}\:{origin}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({b}\right)−\:\:{when}\:{its}\:{coordinate}\:{is}\:{x}. \\ $$$$\:\:\:\:\left(\mathrm{2}\right)−\:{show}\:{that}\:{when}\:{x}\:{is}\:{large}\:{compared}\:{to}\:{distance}\:{d}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{the}\:{force}\:{in}\:\left({b}\right)\:{is}\:{inversely}\:{proportionalto}\:{the} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{cube}\:{of}\:{the}\:{distance}\:{from}\:{origin}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18
![a)Force on 3rd charge by charge A i s repulsive force on 3rd charge by charge B is attrctive net force on 3rd charge when it is at origin is F=(1/(4πε_0 ))(q^2 /d^2 )+(1/(4πε_0 ))(q^2 /d^2 )=2×(1/(4πε_0 ))×(q^2 /d^2 ) direction is −ve yaxis vecorically F_1 ^→ =(1/(4πε_0 ))(q^2 /d^2 )(−j^→ ) F_2 ^→ =(1/(4πε_0 ))(q^2 /d^2 )(−j^→ ) net force=F_1 ^→ +F_2 ^→ =2×(1/(4πε_0 ))(q^2 /d^2 )(−j^→ ) b)F_1 ^→ =(1/(4πε_0 ))(q^2 /((d^2 +x^2 )^(3/2) ))(−dj+xi) F_2 ^→ =(1/(4πε_0 ))(q^2 /((d^2 +x^2 )^(3/2) ))(−dj−xi) F_R ^→ =F_1 ^→ +F_2 ^→ =(1/(4πε_0 ))(q^2 /((d^2 +x^2 )^(3/2) ))(−2dj) net force =(1/(4πε_0 ))(q^2 /((d^2 +x^2 )^(3/2) ))×2d direcection −ve y axis F_R =(1/(4πε_0 ))(q^2 /({x^2 (1+(d^2 /x^2 ))}^(3/2) ))×2d≈(1/(4πε_0 ))×((q^2 ×2d)/x^3 ) [(d/x)→0 as x>>d] [](https://www.tinkutara.com/question/Q46298.png)
$$\left.{a}\right){Force}\:{on}\:\mathrm{3}{rd}\:{charge}\:{by}\:{charge}\:{A}\:{i}\:{s}\:{repulsive} \\ $$$${force}\:{on}\:\mathrm{3}{rd}\:{charge}\:{by}\:{charge}\:{B}\:{is}\:{attrctive} \\ $$$${net}\:{force}\:{on}\:\mathrm{3}{rd}\:{charge}\:{when}\:{it}\:{is}\:{at}\:{origin} \\ $$$${is}\:{F}=\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\frac{{q}^{\mathrm{2}} }{{d}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\frac{{q}^{\mathrm{2}} }{{d}^{\mathrm{2}} }=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }×\frac{{q}^{\mathrm{2}} }{{d}^{\mathrm{2}} } \\ $$$${direction}\:{is}\:−{ve}\:{yaxis} \\ $$$${vecorically} \\ $$$$\overset{\rightarrow} {{F}}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\frac{{q}^{\mathrm{2}} }{{d}^{\mathrm{2}} }\left(−\overset{\rightarrow} {{j}}\right) \\ $$$$\overset{\rightarrow} {{F}}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\frac{{q}^{\mathrm{2}} }{{d}^{\mathrm{2}} }\left(−\overset{\rightarrow} {{j}}\right) \\ $$$${net}\:{force}=\overset{\rightarrow} {{F}}_{\mathrm{1}} +\overset{\rightarrow} {{F}}_{\mathrm{2}} =\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\frac{{q}^{\mathrm{2}} }{{d}^{\mathrm{2}} }\left(−\overset{\rightarrow} {{j}}\right) \\ $$$$\left.{b}\right)\overset{\rightarrow} {{F}}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\frac{{q}^{\mathrm{2}} }{\left({d}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\left(−{dj}+{xi}\right) \\ $$$$\overset{\rightarrow} {{F}}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\frac{{q}^{\mathrm{2}} }{\left({d}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\left(−{dj}−{xi}\right) \\ $$$$\overset{\rightarrow} {{F}}_{{R}} =\overset{\rightarrow} {{F}}_{\mathrm{1}} +\overset{\rightarrow} {{F}}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\frac{{q}^{\mathrm{2}} }{\left({d}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\left(−\mathrm{2}{dj}\right) \\ $$$${net}\:{force}\:=\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\frac{{q}^{\mathrm{2}} }{\left({d}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }×\mathrm{2}{d}\:\:\:{direcection}\:−{ve}\:{y}\:{axis} \\ $$$$ \\ $$$${F}_{{R}} =\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\frac{{q}^{\mathrm{2}} }{\left\{{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{{d}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)\right\}^{\frac{\mathrm{3}}{\mathrm{2}}} }×\mathrm{2}{d}\approx\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }×\frac{{q}^{\mathrm{2}} ×\mathrm{2}{d}}{{x}^{\mathrm{3}} }\:\:\left[\frac{{d}}{{x}}\rightarrow\mathrm{0}\:{as}\:{x}>>{d}\right] \\ $$$$\left[\right. \\ $$$$ \\ $$$$ \\ $$
Commented by Umar last updated on 23/Oct/18
$${thank}\:{you}\:{sir} \\ $$