Question

Electrostatic force between two small spheres at
separation 'r' is 'F'. Now both spheres are touched
and again placed at separation 'r', then electrostatic
Then ratio of
force between them becomes
9F/8
their initial charges is :-
(1) 3:4

(2) 2:1

(3) 3:5

(4)7:2
​

Answer 1

Answer:

Explanation:

Option 3 is the right choice

Answer 2

Answer:

The ratio of their initial charges is 2 : 1.

Explanation:

Let the charges on two spheres be q₁ and q₂ respectively.

The electrostatic force between two small spheres at separation 'r' is 'F' which is equal to:

$F= \frac{kq_1q_2}{r^2}$.....(1)

Now when both the spheres touch each other then the charge on each sphere is q.

where $q = \frac{q_1+q_2}{2}$ (As at the equilibrium charge distribution)

Now, given that the force 9F/8 acts between the spheres which are at the distance 'r'.

$\frac{9F}{8} = \frac{kq^2}{r^2}$

$\implies \frac{9F}{8} = \frac{k(\frac{q_1 + q_2}{2} )^2}{r^2}$........(2)

Divide equation (1) with (2)

$\frac{F}{\frac{9F}{8} } = \frac{\frac{kq_1q_2}{r^2} }{\frac{k(\frac{q_1 + q_2}{2} )^2}{r^2} } \\\implies \frac{8}{9} = \frac{4q_1q_2}{(q_1 + q_2)^2}\\ \implies 2(q_1 + q_2)^2 = 9(q_1q_2)\\\implies 2(q_1^2 +q_2^2 +2q_1q_2) = 9q_1q_2\\\implies 2(\frac{q_1}{q_2} + \frac{q_2}{q_1}+2 ) = 9$......(3)

Let q₁/q₂ = x

Then equation (3) becomes

2(x + (1/x) +2) = 9

⇒ 2x² + 2 + 4x = 9x

⇒ 2x² - 5x + 2 = 0

⇒ 2x² - 4x - x + 2 = 0

⇒ 2x(x - 2) - 1(x - 2) = 0

⇒ (2x - 1)(x - 2) = 0

⇒ x = 1/2 or x = 2/1

Therefore, the ratio of their initial charges is 2 : 1.

#SPJ2