Question

two conducting balls of identical mass m = 10 g and identical charge q hang from nonconducting threads of length L = 130 cm. If x = 3.1 cm, what is q? Since x is much smaller than L approximate sin(?) by ?.

Answer 1

the tension force of the string is T
equilibrium equations:
T*sin theta =F {electrostatic}
T*cos theta =mg*
F=k*q2/(x)2

=>mg*tan theta =k*q2/(x)2

tan theta =sintheta=theta = x/(2L)

=>x^3=k*q^2*2L/mg =>


If L = 130 cm, m = 10 g,and x = 3.1 cm,

q ^2 = x^3*mg/k*2L

= 0.031^3 *10*9.8^2*/9*10^9*2*1.3

=
q=1.22*10^-12 C

Answer 2

from the free body diagram you can write

T cos(theta) = mg------1)

and

T sin(theta) = qE ----2) where E is the field and T is the tension, g is gravitationalconstant.

dividing 1 and 2 we gat

tan(theta) = qE/mg

here tan(theta) = 2x/sqrt(4L^2 - x^2)

calculating we get

q = 7.89 *10^-9 C

Answer 3

So here we imagine two balls of equal mass and charge, so by this we say that each ball feels a force of repulsion (F_repulsion). T, will represent tension in the thread that is non-conducting..

sin(?) = (x/2) / L = (3.1/2) / 130 =====> sin(?) = 0.012

Tcos(?) = mg

F_repulsion = Tsin(?)

= kq^2 / (0.031)^2

From these two bolded equation, we can find tan(?):

tan(?) = (kq2 / (0.031)^2 ) / mg , here we can do a trick because the angle is so small we can substitute tan(?): for sin(?) which we found earlier. Thus:

0.012 = (9x10^2) (q^2) / 0.00096 x 9.81 x 0.01

Solving for q^2:

q^2 = 1.256 x 10^-9 ............. q = 3.54 X 10^-5 C on each ball [ANS]

Hope this helps!!