Question

Q.Two identical charges each of charge q are placed at a distance 2l, a third charge Q is placed at a centre of line joining both charges. If we slightly displaced 3rd charge along the line joining the two charges then find out the time period of oscillation.

Answer 1

Given:

Two identical charges each of charge q, placed at a distance 2l.

A third charge Q is placed at a centre of line joining both charges.

To Find :

The time period of oscillation, If we slightly displaced 3rd charge along the line joining the two charges.

Solution:

General form of a Simple Harmonic Motion is given by,

• $\frac{d^{2}x }{dt^{2} }$ + ω²x = 0,
• Where ω is the frequency of oscillation.

Consider the charge kept at the center and is slightly displaced by a distance 'x'.

Then Q will be at a distance ol l+x from one of the 'q' charges and l - x from the other.

Then net Force on Q is given by,

F = Force due to q at l + x - Force due to q at l - x

• F = kQq/(l + x)² - KQq/(l - x)²
• F = kQq( 1/(l+x)² - 1/(l-x)²)

Taking l² outside,

• F = KQq/l² ( $(1+ \frac{x}{l} )^{-2}$ - $(1- \frac{x}{l})^{-2}$)

Applying binomial expansion for x<<l ,

• F = KQq/l² (( 1 -  $\frac{2x}{l}$ ) - (1 - $\frac{-2x}{y}$))
• F = KQq/l²( 1 - 2x/l - 1 - 2x/l)
• F = KQq/l² ( -4x/l)

If mass of the charge particle = m ,

• m a = 4KQq/l³ ( -x )
• a = d²x/dt² = (- x ) 4kQq/ml³

Therefore, from the general equation of SHM,

• ω² = 4KQq/ml³
• ω = 2$\sqrt{\frac{KQq}{ml^{3} } }$

Therefore time period of oscillation is,

• T = 2π/ω
• T = π$\sqrt{\frac{ml^{3} }{KQq} }$  , where K = $\frac{1}{2\pi\epsilon_0}$

The time period of oscillation, is T = π$\sqrt{\frac{ml^{3} }{KQq} }$