Question

Two balls of same mass and carrying equal charge are hung from a fixed support of length l. At electrostatic equilibrium, assuming that angles made by each thread is small, the separation, x between the balls is proportional to:

Answer 1

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Answer 2

The separation between the balls is proportional to $x\propto (l)^{\dfrac{1}{3}}$

Explanation:

Given that,

Two balls of same mass and carrying equal charge are hung from a fixed support of length l.

According to diagram

We need to calculate the tension

Using balance equation

$T\cos\theta=mg$.....(I)

$T\sin\theta=\dfrac{kq^2}{x^2}$.....(II)

Divided equation (II) by equation (I)

$\tan\theta=\dfrac{kq^2}{mgx^2}$....(III)

We need to calculate the proportional term

Using diagram,

$\tan\theta=\dfrac{CB}{OB}$

$\tan\theta=\dfrac{\dfrac{x}{2}}{(l^2-\dfrac{x^2}{4})^{\frac{1}{2}}}$

Put the value from equation (III)

$\dfrac{\dfrac{x}{2}}{(l^2-\dfrac{x^2}{4})^{\frac{1}{2}}}=\dfrac{kq^2}{mgx^2}$

$x^3=\dfrac{2kq^2}{mg}(l^2-\dfrac{x^2}{4})^{\frac{1}{2}}$

If is small then, $\dfrac{x^2}{4}<<l^2$

$x^3=\dfrac{2kq^2}{mg}(l^2)^{\frac{1}{2}}$

$x=\dfrac{2kq^2}{mg}(l)^{\frac{1}{3}}$

$x\propto l^{\frac{1}{3}}$

Hence, The separation between the balls is proportional to $x\propto (l)^{\frac{1}{3}}$

Learn more :

Topic : electrostatic

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