Point charge are placed at the verticas of a squre of side a as shown in fig. What should be the sign of charge q and magnitude of the ratio `|q//Q|` so that
(i) net force on each Q is zero?
(ii) net force on each q is zero?
Is it possible that the entire system could be in electrostatic equilibrium?

Options

A. `(1)/(2 sqrt(2))` `20 sqrt2``
B. `(1)/(2 sqrt(2))` `2 sqrt2`
C. `(1)/(2 sqrt(2))` `12 sqrt2`
D. `(1)/(2 sqrt(12))` `2 sqrt2`

Answer

B

Explanation

Consider the force action on charge Q placed at (shown in fig. )
Case I. Let the charge q and Q be of same sign.
here,
`(F_1)=1/(4 pi epsilon_(0)) (qQ)/(a^2)` (force of q at D on Q at A)
,
`(F_2)=1/(4 pi epsilon_(0)) (qQ)/(a^2)` (force of q atB on Q at A)
`(F_3)=1/(4 pi epsilon_(0)) (qQ)/(a^2)` (force of q atC on Q at A)
In fig. the resultant of force `vec(F_1)` and `vec(F_2)` will along `vec(F_3)` so that the net force on Q can not be zero. Hence, q and Q have to be of opposite signs.
Case II. Let the charges q and Q be of opposite signs. In this case, as shown in fig. resultant of `vec(F_1)` and `vec(F_2)` will be opposite to `vec(F_3)` so that it bexomes possible to obtain a condition of zero net force. Let us write `vecF_(R ) =vecF_(1) + vecF_(2)`. so
`sqrt(F_(1)^(2)+F_(2)^(2)) = (1)/(4 pi epsilon_(0)) (qQ)/(a^(2))sqrt(2)`
The direction of `vecF_(R )` will be along AC (`vecF_(R )`, being the resultant of force of equal magnitude, bisects the angle between the two). `vecF_(R )` and `vecF_(3)` are omn opposite directions. Theat force on Q can be zero if their magnitudes are also equal. i,e.
`(1)/(4 pi epsilon_(0)) (qQ)/(a^(2))sqrt(2)=(1)/(4 pi epsilon_(0)) (Q Q)/(2a^(2))` or `(Q)/(4 pi epsilon_(0))(q sqrt(2)-(Q)/(2))=0`
or `q=(Q)/(2 sqrt(2))` or `|(q)/(Q)|=(1)/(2 sqrt(2)) {Q!=0}`
Therefore, the sign of q should be the negative of Q.
(ii) Consider now the forces acting on charge q placed at B. In a similar manner, as dicussed in (i) , for the net force on q to be zero. q and Q have to be of opposite signs. This is also shown in the given figure.
`F_(1)=(1)/(4 pi epsilon_(0)) (qQ)/(a^(2))` (force of Q at A on q at B)
`F_(2)=(1)/(4 pi epsilon_(0)) (qQ)/(a^(2))` (force of Q at C on q at B)
`F_(3)=(1)/(4 pi epsilon_(0)) (qQ)/(a^(2))` (force of q at D on qat B)
,
From fig. we can write
`vecF_(R ) =vecF_(1) + vecF_(2)`
`:. F_(R) =sqrt(F_(1)^(2)+F_(2)^(2))=(1)/(4 pi epsilon_(0)) (Qq)/(a^(2)) sqrt(2)`
The resultant of `vecF_(1)` and `vecF_(2)` i.e `vecF_(R)` is opposite to `vecF_(3)` . Net force can become zero if their magnitudes are also equal i.e.
`(1)/(4 pi epsilon_(0)) (Qq)/(a^(2))sqrt(2) =(1)/(4 pi epsilon_(0)) (q^(2))/(2a^(2))` or `(q)/(4 pi epsilon_(0)a^(2)) [sqrt(2)Q-(q)/(2)]=0`
or `Q=(q)/(2 sqrt(2))` or `|(q)/(Q)| = 2 sqrt(2) {q!=0}`
Therefore, the sign of q should be negative of Q. In this case, we need not to repeat the calculation as the present situation id same as the previous one, we can directly write `|q//Q|=2 sqrt(2)`. The entire system cannot be in equilibrium since both conditions i.e. `q=-Q//2 sqrt(2)` and `Q=-q//2sqrt(2)` cannot be satisfied together.