Question

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.​

Answer 1

$\mathfrak{\underline{\underline{\red{Question: }}}}$

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.

$\mathfrak{\underline{\underline{\purple{Solution↑↑↑: }}}}$

Answer 2

Given :

• Q ₁  = 3 x 10-¹⁹ C
• Q ₂ = 12 x 10-19 C
• Distance between the two charges = 2.5 cm

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 ,

$\implies E_1 - E_2 = 0$

$\implies E_1 = E_2$

$\implies \dfrac{kQ_1}{r^2} = \dfrac{kQ_2}{r'^2}$

$\implies \dfrac{3 \times 10^{-19}}{x^2} = \dfrac{12 \times 10^{-19}}{(2.5-x) ^2}$

$\implies \dfrac{1 }{x^2} = \dfrac{4 }{(2.5-x) ^2}$

Taking square root on both the sides we get,

$\implies \dfrac{1 }{x} = \dfrac{2 }{2.5-x}$

$\implies 2.5 - x = 2x$

$\implies 3x = 2.5$

$\implies x = \dfrac{2.5}{3}$

$\implies \boxed{x = 0.83 m }$

The electric field intensity is zero at a distance of 0.83 m from Q ₁ and 1.67 m from Q ₂