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Two equal negative charges -q are fixed ...

Two equal negative charges `-q` are fixed at points `(0, -a)` and `(0,a)` on y-axis. A poistive charge Q is released from rest at point `(2a, 0)` on the x-axis. The charge Q will

A

`sqrt((2q^2)/(4pi epsilon_0 ma^3))`

B

`sqrt((2q^2)/(2pi epsilon_0 ma^3))`

C

`sqrt((q^2)/(2pi epsilon_0 ma^3))`

D

`sqrt((q^2)/(pi epsilon_0 ma^3))`

Text Solution

Verified by Experts

The correct Answer is:
C
Restoring force on charge q –
`F_("net") =-2F cos theta`
`=(-2Kq^2)/((x^2+a^2)).x/((a^2+x^2)^(1//2))`
`= (-2Kq^2x)/((a^2+x^2)^(3//2))`

As `a gt gt x , F_("net") = (-2Kq^2x)/(a^3)`
Acceleration (A)
`(-2Kq^2x)/(ma^3)`
For linear S.H.M. –
`A =-omega^2x`
`omega^2 = (2Kq^2)/(ma^3)`
`omega=sqrt((2q^2)/(4piepsilon_0ma^3))`
`omega=sqrt((q^2)/(2piepsilon_0ma^3))`
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