Menu Close

Question-131199

Tinku Tara
Pin




Question Number 131199 by aurpeyz last updated on 02/Feb/21
Answered by physicstutes last updated on 02/Feb/21
Example 4 Q_1 = 2.0 μC and Q_2 = −4.0 μC, R = 50 cm the neutral point lies a distance of x from Q_1 and (0.5−x) m from Q_2 We define the electric field strenght as E = (F/Q) E_Q_1 = ((kQ_2 )/R_1 ^2 ) = ((k(4.0 μC))/x^2 ) E_(Q_2 ) = ((kQ_1 )/R_2 ^2 ) = ((k(2.0 μC))/((0.5−x)^2 )) at the neutral point E_Q_1 = E_Q_2 ⇒ ((4.0)/x^2 ) = ((2.0)/((0.5−x)^2 )) ⇒ (2/x) = ((√2)/(0.5−x)) 1−2x = (√2) x ⇒ x = (1/(2+(√2))) ≈ 0.3 m or 30 cm from Q_1
$$\boldsymbol{\mathrm{Example}}\:\mathrm{4} \\ $$$${Q}_{\mathrm{1}} \:=\:\mathrm{2}.\mathrm{0}\:\mu\mathrm{C}\:\mathrm{and}\:{Q}_{\mathrm{2}} \:=\:−\mathrm{4}.\mathrm{0}\:\mu\mathrm{C},\:{R}\:=\:\mathrm{50}\:\mathrm{cm} \\ $$$$\mathrm{the}\:\mathrm{neutral}\:\mathrm{point}\:\mathrm{lies}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{of}\:{x}\:\mathrm{from}\:{Q}_{\mathrm{1}} \:\mathrm{and}\:\left(\mathrm{0}.\mathrm{5}−{x}\right)\:\mathrm{m}\:\mathrm{from}\:{Q}_{\mathrm{2}} \\ $$$$\mathrm{We}\:\mathrm{define}\:\mathrm{the}\:\mathrm{electric}\:\mathrm{field}\:\mathrm{strenght}\:\mathrm{as}\:\boldsymbol{\mathrm{E}}\:=\:\frac{\boldsymbol{\mathrm{F}}}{{Q}} \\ $$$$\:\mathrm{E}_{{Q}_{\mathrm{1}} } \:=\:\frac{{kQ}_{\mathrm{2}} }{{R}_{\mathrm{1}} ^{\mathrm{2}} }\:=\:\frac{{k}\left(\mathrm{4}.\mathrm{0}\:\mu\mathrm{C}\right)}{{x}^{\mathrm{2}} } \\ $$$$\:\mathrm{E}_{{Q}_{\mathrm{2}} \:} =\:\frac{{kQ}_{\mathrm{1}} }{{R}_{\mathrm{2}} ^{\mathrm{2}} }\:=\:\frac{{k}\left(\mathrm{2}.\mathrm{0}\:\mu\mathrm{C}\right)}{\left(\mathrm{0}.\mathrm{5}−{x}\right)^{\mathrm{2}} } \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{neutral}\:\mathrm{point}\:\:{E}_{{Q}_{\mathrm{1}} } \:=\:{E}_{{Q}_{\mathrm{2}} } \\ $$$$\Rightarrow\:\frac{\mathrm{4}.\mathrm{0}}{{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{2}.\mathrm{0}}{\left(\mathrm{0}.\mathrm{5}−{x}\right)^{\mathrm{2}} }\:\:\Rightarrow\:\frac{\mathrm{2}}{{x}}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{0}.\mathrm{5}−{x}} \\ $$$$\:\mathrm{1}−\mathrm{2}{x}\:=\:\sqrt{\mathrm{2}}\:{x} \\ $$$$\Rightarrow\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:\approx\:\mathrm{0}.\mathrm{3}\:\mathrm{m}\:\mathrm{or}\:\mathrm{30}\:\mathrm{cm}\:\mathrm{from}\:\mathrm{Q}_{\mathrm{1}} \\ $$
Commented by aurpeyz last updated on 06/Feb/21
Pls can you explain the neutral point
$${Pls}\:{can}\:{you}\:{explain}\:{the}\:{neutral}\:{point} \\ $$$$ \\ $$
Commented by physicstutes last updated on 06/Feb/21
point were E_(net) = 0
$$\mathrm{point}\:\mathrm{were}\:{E}_{{net}} \:=\:\mathrm{0} \\ $$

Pin