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A charge q is placed at the centre of th...

A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to:

A

`-(Q)/(2)`

B

`-(Q)/(4)`

C

`-4Q`

D

`+(Q)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
B

`F = (kQq)/((r)/(2))^(2) +(kQ Q)/(r^(2)) = 0`
`q =- (Q)/(4)`
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Knowledge Check

  • If a charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium then the value of q is

    A
    (a) `Q/2`
    B
    (b) `-Q/2`
    C
    (c) `Q/4`
    D
    (d) `-Q/4`

  • A charge q is placed at the center of the line joining two equal charges Q . Three charges are in equilibrium then what will be the value of 'q' ?

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    `(Q)/(4)`
    B
    `(-Q)/(4)`
    C
    Q
    D
    `(-Q)/(8)`

  • Two point charges q_(1) = +2 C and q_(2) = - 1C are separated by a distance d . The position on the line joining the two charges where a third charge q = + 1 C will be in equilibrium is at a distance

    A
    `(d)/(sqrt(2))` from `q_(1)` and `q_(2)`
    B
    `(d)/(sqrt(2))` from `q_(1)` away from `q_(2)`
    C
    `(d)/(sqrt(2) - 1)` from `q_(2)` between `q_(1)` and `q_(2)`
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