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Three charges Q, +q and +q are placed at...

Three charges Q, `+q` and `+q` are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if Q is equal to

A

`(-q)/(1+sqrt(2))`

B

`(-2q)/(2+sqrt(2))`

C

`-2q`

D

`+q`

Text Solution

Verified by Experts

(b) The net eelctrostatic enrgy is
`U= (KQq)/a = (KQq)/( sqrt 2a) + (Kqq)/a =0`
`Q+ Q/(sqrt2)+q =0`

which gives ` Q =-q ((sqrt2)/(sqrt 2 + 1) = (-2 q)/(2 + sqrt2)`
(c )
`E_1 ` is field due to ` 8 mu C` and ` 8mu C` at `P` will be in the direction of ` OP` while `vec E_2` due to `-1 mu C` and `-1 mu C` will be in the direcition of `PO` For `E_p =0` at `P (OP = x)`
` 2 E_1 cs alpha = 2 E_2 cos theta`
`(2 K xx 8 xx 10^(-6))/( (sqrt(27 3//2)^2 + x^2) xx x/((sqrt(3 //2))^2 +x^2)^(1//2)`
`x = -+ (sqrt 5)/2`
For `OP gt x, E_p` will be towards `OP` and for `OP lt x, E_p` will be tward ` OP` for ` E_p=O` From cosnervation of energy
(i) For least value of `V` to cross origing
Loss of `K. e. = `Gain of `P.E.`
`1/2 mv_0^2 = q V_p` ....(1)
` v+p = 2 xx 9 xx10^9 [(8 xx 10^(-4))/(sqrt ((sqrt (27)/2)^2 + (sqrt 5/2)^2) sqrt) - (1 xx 1-^(-6))/(sqrt ((sqrt 3/2)^2) +( sqt 5/2))^2]`
` V_p = 27000 ` volt ...(2)
Form (1) and (2)
`1/2 xx 6 xx 10^(-4) xx v_0^2 = 0. ^(-6) xx 27 00`
`v_0^2 =9`
`v_0 = 3 m//s`
(ii) For `KE` at origin
Potential at origin `v_0 = 10^(-7) xx 2 xx 9 xx 10^9`
` [( 8 xx 10^(-6))/(sqrt 27 //2) - (1 xx 10^(-6))/( sqrt 3//2)] = 2 . 4 xx 10^4 ` volt
` KE` at origin`= q ( V_0 - V_p ) =- 0.1 xx 10^(-6)`
`(24000 - 27000) = 3 xx 10^(-4) J`.
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