Three 3.0 g balls are tied to 80-cm-long threads and hung from a single fixed point. Each of the balls is given the same charge q. At equilibrium, the three balls form an equilateral triangle in a horizontal plane with 20 cm sides. What is q?

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

ok

step-by-step explanation:

ou would have 12 of each