Answer:
q = 0.105uC
Step-by-step explanation:
We can determine the force on one ball by assuming two balls are stationary, finding the E field at the lower right vertex and calculate q from that.
Considering the horizontal and vertical components.
First find the directions of the fields at the lower right vertex. From the lower left vertex the field will be at 0° and from the top vertex, the field will be at -60° or 300° because + charge fields point radially outward in all directions. The distances from both charges are the same since this is an equilateral triangle. The fields have the same magnitude:
E=kq/r²
Where r = 20cm
= 20/100
= 0.2m
K = 9.0×10^9
9.0×10^9 × q /0.2²
9.0×10^9/0.04
2.25×10^11 q
These are vector fields of course
Sum the horizontal components
Ecos0 + Ecos300 = E+0.5E
= 1.5E
Sum the vertical components
Esin0 + Esin300 = -E√3/2
Resultant = √3E at -30° or 330°
So the force on q at the lower right corner is q√3×E
The balls have two forces, horizontal = √3×E×q
and vertical = mg, therefore if θ is the angle the string makes with the vertical tanθ = q√3E/mg
mg×tanθ = q√3E.
..1
Then θ will be...
Since the hypotenuse = 80cm
80cm/100
= 0.8m
The distance from the centroid to the lower right vertex is 0.1/cos30 =
0.1/0.866
= 0.1155m
Hence,
0.8×sinθ = 0.1155
Sinθ = 0.1155/0.8
Sin θ = 0.144375
θ = arch sin 0.144375
θ = 8.3°
From equation 1
mg×tanθ = q√3E
g = 9.8m/s^2
m = 3.0g = 0.003kg
0.003×9.8×tan(8.3)
0.00428 = q√3E
0.00428 = q×1.7320×E
Where E=kq/r²
Where r = 0.2m
0.0428 = kq^2/r² × 1.7320
K = 9.0×10^9
0.0428/1.7320 = 9.0×10^9 × q² / 0.2²
0.02471×0.04 = 9.0×10^9 × q²
0.0009884 = 9.0×10^9 × q²
0.0009884/9.0×10^9 = q²
q² = 109822.223
q = √109822.223
q = 0.105uC
