Question

Q= If two point charges +9e and +1e are kept at a distance of 16 cm from each other, then at what point between these charges should a third charge q be placed so that it remains in equilibrium____. PLZZ SOLVE IT..

Answer 1

hope this will help u....

Answer 2

Answer:

The distance of the third charge from the first charge is 0.04 m

Explanation:

The charge on the first particle is +9e

The charge on the second particle is +e

The distance between both the particle is L=0.16 m

Let the distance x from the first particle to which the third particle of charge q will be in equilibrium

The electric force between first and third charge is

$F_{1}=\dfrac{9keq}{x^2}$

The electric force between the second and third particle is

$F_{2}=\dfrac{keq}{(L-x)^2}$

As the net force on the third charge is zero

So equating both the forces

$F_{1}=F_{2}\\\dfrac{9keq}{x^2}=\dfrac{keq}{(L-x)^2}\\\dfrac{9}{x^2}=\dfrac{1}{(L-x)^2}\\(L-x)^2=9x^2$

Taking square root at both sides we get,

$\sqrt{(L-x)^2}=\sqrt{9x^2}\\L-x=3x\\4x=L\\4x=0.16\\x=\dfrac{0.16}{4}\\x=0.04\ m$