Question

Two fixed charges +1 microC each are 4 cm as apart. A third charge of -1 microC and mass 1g is placed at the centre and given a small displacement along the perpendicular bisector. Find the time period of oscillation of the ensuing SHM in millisecond.

Answer 1

time period of oscillation of the ensuing SHM in millisecond is 3.

two charges each of magnitude +q are placed at a fixed distance L apart . another charge +Q and mass m is placed at midpoint of the line joining the charges . Let charge +Q displaces x slightly along A.

A(+q).....(L/2)........(+Q).........(L/2).......B(+q)

when charge +Q displaces x slightly along A.

force experienced by +Q due to charge A , F = KqQ/(L/2 - x)²

and force experienced by +Q due to charge B = KqQ/(L/2 + x)²

now, net force experienced by +Q = KqQ/(L/2 - x)² - KqQ/(L/2 + x)²

= KqQ[ {(L/2 + x)² - (L/2 - x)²}/(L²/4 - x²)² ]

= KqQ[L(2x)/(L²/4 - x²)]

here, L >> x

so, L²/4 >> x² and L²/4 - x² ≈ L²/4

Fnet = 2KqQLx/(L²/4)² = 32KqQx/L³

here it is clear that, force is directly proportional to x and direction of force is just opposite to direction of displacement. [ as net force acting along B direction but displacement is along A ]

so, we can use, Fnet = mω²x

⇒32KqQ/L³ = mω²

⇒ω² = 32KqQ/mL³

⇒ω = √{32KqQ/mL³}

⇒2π/T = √{32KqQ/mL³}

⇒T = 2π√{mL³/32KqQ}

here, m = 10¯³ kg, K = 9 × 10^9 Nm²/C², L = 4 × 10¯²m , q = 10^-6 C , Q = 10^-6 C .

T = 2π√{ 10^-3 × (4 × 10^-2)³/32 × 9 × 10^9 × 10^-6 × 10^-6}

= 2π√{64 × 10^10^-9/288 × 10^-3}

= 2π√{2/9 × 10^-6}

= 2√2π/3 × 10¯³ sec

≈ 3 × 10¯³ sec ≈ 3 ms

time period of oscillation of the ensuing SHM in millisecond is 3