Answer:
Third charge should be placed at a distance d/3
Sign is negative
Magnitude is |q| = 4Q/3
Explanation:
To find a place where a third charge, q, is placed so that all three charges will be in equilibrium would be at;
E1 = E2
Now, let the distance between the charges +Q and +4Q be d. Also, let the distance from charge +Q to the third charge q be x.
Thus;
E1 = kQ/x²
E2 = k(4Q)/(d - x)²
Since equilibrium, then;
E1 = E2 gives;
kQ/x² = k(4Q)/(d - x)²
k and Q will cancel out to give;
1/x² = 4/(d - x)²
Cross multiply to get;
(d - x)² = 4x²
Taking square root of both sides gives;
√(d - x)² = √4x²
Gives;
d - x = 2x
d = x + 2x
d = 3x
x = d/3
This is the point at which E1 = E2.
Since at equilibrium, it means that Fq + F2 = 0
Now, electric force formula is;
F = kq1*q2/r²
So; Fq = kqQ/x²
F2 = kQ(4Q)/d²
Equating Fq + F2 = 0 gives;
kqQ/x² + kQ(4Q)/d² = 0
kqQ/x² = -kQ(4Q)/d²
Like terms cancel out to get;
q/x² = -4Q/d²
q = - 4Qx²/d²
earlier we saw that x = d/3.
So, q = -4Q × (d/3)²/d²
q = -4Q × ⅓
q = - 4Q/3
Thus,it's sign is negative.
It's magnitude will be the positive value.
Thus, magnitude is;
|q| = 4Q/3
