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

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graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.

answer: yes, since the line has a steeper rate of change than the exponential the y-values will be greater and they will only intersect one time.