Four positive charges `(2sqrt2-1)Q` are arranged at the four corners of a square. Another charge q is placed at the center of the square. Resulting force acting on each corner charge is zero if q is

Options

A. `-(7Q)/4`
B. `-(4Q)/7`
C. `-Q`
D. `-(sqrt2+1)Q`

Answer

A

Explanation

`..`
`Q_1=(2sqrt2-1)Q`
`F_1` = force between `Q_1 and Q_1` at distance a
`F_2=Force between `Q_1 and Q_1` at distance `sqrt2a`
`F_3=` Force between `Q_1` and q at between `a/sqrt2`
for `F_3` to be in the shown direction, q and `Q_1` should have opposite signs. For net charge to be zero on `Q_1` placed at `P`.
`|F_3|=`Resultant of `F_1, F_2 and F_1`
`:. (kQ_1q)/(a//sqrt2)^2=sqrt2F_1+F_2`
`=sqrt2[(kQ_1Q_2)/a^2]+(kQ_1Q_1)/(sqrt2a)^2`
`:. |q|=Q_1/2 [sqrt2+1/2]=((2sqrt2+1)/4)Q_1`
`=((2sqrt2+1)/4)(2sqrt2-1)Q`
`=7/4 Q` or `q=-7/4Q ltbr