Two point charges of +3×10-19C and +12×10-19C are separated by a distance of 2.5m.Find the point on the line joining the mat which the electricfield intensity is zero.
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Two point charges of +3×10-19C and +12×10-19C are separated by a distance of 2.5m.Find the point on the line joining the mat which the
[tex]\mathfrak{\underline{\underline{\red{Question: }}}}[/tex]
twσ pσínt chαrgєѕ σf +3×10-19C αnd +12×10-19C αrє ѕєpαrαtєd вч α distancє of 2.5m. fínd thє pσínt σn thє línє jσíníng thє mαt whích thє electríc fíєld íntєnѕítч is zerσ.
[tex]\mathfrak{\underline{\underline{\purple{Solution↑↑↑: }}}}[/tex]
Given :
Solution :
As per the given data,
There is a point on the line joining the two charges where the electric field intensity is zero.
Let, the distance of the point from Q ₁ be x and from Q ₂ be 2.5 - x
Now ,
[tex]\implies E_1 - E_2 = 0 \\ \\[/tex]
[tex]\implies E_1 = E_2 \\ \\[/tex]
[tex]\implies \dfrac{kQ_1}{r^2} = \dfrac{kQ_2}{r'^2} \\ \\[/tex]
[tex]\implies \dfrac{3 \times 10^{-19}}{x^2} = \dfrac{12 \times 10^{-19}}{(2.5-x) ^2} \\ \\[/tex]
[tex]\implies \dfrac{1 }{x^2} = \dfrac{4 }{(2.5-x) ^2} \\ \\[/tex]
Taking square root on both the sides we get,
[tex]\implies \dfrac{1 }{x} = \dfrac{2 }{2.5-x} \\ \\[/tex]
[tex]\implies 2.5 - x = 2x \\ \\[/tex]
[tex]\implies 3x = 2.5 \\ \\[/tex]
[tex]\implies x = \dfrac{2.5}{3} \\ \\[/tex]
[tex]\implies \boxed{x = 0.83 m } \\ \\[/tex]
The electric field intensity is zero at a distance of 0.83 m from Q ₁ and 1.67 m from Q ₂