Calculating Tension and Charge Magnitude in Charged Particle Systems

[MrBballa]

Homework Statement

Hey guys I was wondering if you can help me out on these two questions i have. They're due by Wednsday so i'd appreciate if i can get a reply to se if my work is right on these by then. Thanks

Question 1)
A small spherical insulator of mass 80 grams and charge +1.80 x 10 ^-6 ( Mew) C is hung by a thin massless wire. A charge of -2.00 x 10 ^-6 (Mew) C is held 0.20 meters away from the
sphere and directly to the right of it, so the wire makes an angle (Theta) with the vertical. Find the angle and the tension in the wire.

Question 2) Two identical point charges of +2.00 x 10 ^-6 (Mew) C are fixed to diagonally opposite corners of a square. A third charge is then moved to and fixed at the center of the square, such that it causes the potentials at the empty corners of the square to change signs without changing magnitudes. Find the sign and magnitude of the third charge.

Homework Equations

Question 1)
Let T be the tension in the wire.
A vertical force balance of the insulator says:
T sin theta = M g
A horizontal force balance says :
T cos theta = k Q1*Q2/R^2 which is Coulomb's law. k is the Coulomb constant and Q1 and Q2 are the two charges.
Solve the two equations in the two unknowns, theta and T.
Divide one equation by the other to get tan theta and eliminiate T.
Question 2)
Let the charges at the two opposite corners be q, and the added charge at the center be q'. Before q' is added, the potential at the empty corners is
2 k q/a, where a is the length of a side of the square.
The distance of the center of the square from either empty corner is a/sqrt2
The potential at the emptry corners with q' added must be -2kq/a = 2kq/a + kq' sqrt2/a
The k's and a's cancel and you are left with -2q = +q' *sqrt2
q' = -sqrt2*q (since 2/sqrt2 = sqrt2)

The Attempt at a Solution

Question 1)
M= 80 grams =.08 kg... g= 9.8m/s^2
Q1= 1.8 x 10 ^-6... Q2= 2 x 10 ^-6... r= .2 meters
Tan (theta) = k (Q1 x Q2)/r^2
Theta= 44.07
Plug Theta into Vertical Force equation: Tsin(44.07)= (.08)(9.8)... T=1.13
Question 2
q= 2 x 10^-6
Find q': q' = -sqrt2*(2 x 10^-6)
q'= - 2.83 x 10 ^-6 C

Let me know if these answers are right or if i need to correct something in my calculations Thanks again for the help.

 

[supratim1]

i think you are right.

 

[supratim1]

and Welcome to PF!