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Question Number 13049 by tawa tawa last updated on 12/May/17

Two charges q_1 = 10 μC and q_2 = 5 μC are placed on the axis at A (10, 0) cm and (20, 0) cm respectively. Determine a position between the two charges where the electric field intensity is 0.

$$\mathrm{Two}\:\mathrm{charges}\:\mathrm{q}_{\mathrm{1}} \:=\:\mathrm{10}\:\mu\mathrm{C}\:\mathrm{and}\:\:\mathrm{q}_{\mathrm{2}} \:=\:\mathrm{5}\:\mu\mathrm{C}\:\:\mathrm{are}\:\mathrm{placed}\:\mathrm{on}\:\mathrm{the}\:\mathrm{axis}\:\mathrm{at}\:\mathrm{A}\:\left(\mathrm{10},\:\mathrm{0}\right)\:\mathrm{cm} \\ $$$$\mathrm{and}\:\left(\mathrm{20},\:\mathrm{0}\right)\:\mathrm{cm}\:\:\mathrm{respectively}.\:\mathrm{Determine}\:\mathrm{a}\:\mathrm{position}\:\mathrm{between}\:\mathrm{the}\:\mathrm{two}\:\mathrm{charges} \\ $$$$\mathrm{where}\:\mathrm{the}\:\mathrm{electric}\:\mathrm{field}\:\mathrm{intensity}\:\mathrm{is}\:\mathrm{0}.\: \\ $$

Answered by ajfour last updated on 12/May/17

say it is so at x. a=10cm=0.1m , q_1 =10μC b=20cm=0.2m, q_2 =5μC for zero intensity of electric field at (x,0) : (q_1 /(4πε_0 (x−a)^2 )) =(q_2 /(4πε_0 (b−x)^2 )) ⇒ ((b−x)/(x−a)) =(√(q_2 /q_1 )) = (1/(√2)) b(√2) −x(√2) =x−a x=((a+b(√2))/(1+(√2))) =(a+b(√2) )((√2)−1) x=(0.1+0.2(√2))((√2)−1) x=(((1+2(√2)))((√2)−1))/(10)) m x=((3−(√2))/(10)) m = 0.1586 m electric field intensity is zero at (15.86, 0) cm .

$${say}\:{it}\:{is}\:{so}\:{at}\:{x}. \\ $$$${a}=\mathrm{10}{cm}=\mathrm{0}.\mathrm{1}{m}\:,\:{q}_{\mathrm{1}} =\mathrm{10}\mu{C} \\ $$$${b}=\mathrm{20}{cm}=\mathrm{0}.\mathrm{2}{m},\:\:\:{q}_{\mathrm{2}} =\mathrm{5}\mu{C} \\ $$$${for}\:{zero}\:{intensity}\:{of}\:{electric}\:{field} \\ $$$${at}\:\:\left({x},\mathrm{0}\right)\:: \\ $$$$\frac{{q}_{\mathrm{1}} }{\mathrm{4}\pi\epsilon_{\mathrm{0}} \left({x}−{a}\right)^{\mathrm{2}} }\:=\frac{{q}_{\mathrm{2}} }{\mathrm{4}\pi\epsilon_{\mathrm{0}} \left({b}−{x}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\frac{{b}−{x}}{{x}−{a}}\:=\sqrt{\frac{{q}_{\mathrm{2}} }{{q}_{\mathrm{1}} }}\:\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\: \\ $$$${b}\sqrt{\mathrm{2}}\:−{x}\sqrt{\mathrm{2}}\:={x}−{a} \\ $$$${x}=\frac{{a}+{b}\sqrt{\mathrm{2}}}{\mathrm{1}+\sqrt{\mathrm{2}}}\:=\left({a}+{b}\sqrt{\mathrm{2}}\:\right)\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$${x}=\left(\mathrm{0}.\mathrm{1}+\mathrm{0}.\mathrm{2}\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$${x}=\frac{\left.\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\right)\right)\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{10}}\:{m} \\ $$$${x}=\frac{\mathrm{3}−\sqrt{\mathrm{2}}}{\mathrm{10}}\:{m}\:=\:\mathrm{0}.\mathrm{1586}\:{m} \\ $$$${electric}\:{field}\:{intensity}\:{is}\:{zero}\:{at} \\ $$$$\:\:\:\left(\mathrm{15}.\mathrm{86},\:\mathrm{0}\right)\:{cm}\:. \\ $$

Commented by tawa tawa last updated on 12/May/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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