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

2 and 1.

step-by-step explanation:

plug in each answer. 0 doesn't work, 1 does. so does 2. 4 doesn't:

0 = 1 is not true.

2 = 2 is true.

4 = 4 is true.

8 = 16 is not true.

hope this !

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3