Question

Two charges, QL and QR have the same magnitude of charge. QR is fixed in place as shown in the diagram below. A second charge QL is attached to a massless string which makes an angle of 22.1° with respect to the horizontal. QL is not moving (and not accelerating) and is located d = 0.317 m directly to the left of QR. QL has a mass of 12.5 kg.

Answer 1

What is the magnitude of the charge on QL (in μC)?

Answer 2

180-22.1=157.9
9.8*12.5/sin(157.9)=325.6
325.6*cos(157.9)=-301.68

Answer 3

Q=sqrt(301.68)*(.317)^2/k=3372138100

Answer 4

Do i then times by 10^6?

Answer 5

Since the charge QL is not moving, you need to balance the forces acting on it.

Answer 6

Im not sure I understand

Answer 7

Components of T are as

Answer 8

Since QL is not moving
\(F_g = T sin \theta\) and
\(F = T cos \theta\)

Answer 9

T is not given. So eliminate T by dividing these two equations and you get
\(\frac{F_g}{F} = \frac{sin \theta}{cos \theta }\)

Answer 10

so sin(22.1)/cos(22/1)?

Answer 11

use calculator for this

Answer 12

yep I got .406

Answer 13

Right!

Answer 14

its due in four minutes lol no pressure but could you help me with the rest?

Answer 15

Let the charge on QL (or QR) is q, then \( F = k\frac{q^2}{d^2}\) and \(F_g = mg = 12.5 g\)
\(\frac{sin 22.1^o}{cos 22.1^o} = tan 22.1^o = 0.406\)
Calculate q.

Answer 16

wait so solve for what?

Answer 17

Solve for q. Use k = 9*10^9

Answer 18

whats F?

Answer 19

Oh crap times gonna run out