- #1
mlostrac
- 83
- 0
Homework Statement
How much energy is need to place four positive charges, each of magnitude +5.0 mC, at the vertices of a square of side 2.5 cm?
Homework Equations
E = kQ/r^2
The Attempt at a Solution
I'm really not sure if what I did is close to correct, but I got an answer.
1) I calculated the Force experienced by 1 charge by another (arbitrarily picked):
F = k Q^2/d^2
= 3.6 x 10^8
Therefore, if I picked the bottom left charge of the square, it is experiencing a repulsive force in the -y and -x directions each of 3.6 x 10^8 N.
2) The third charge, the upper right, was also taken into account after finding the distance from the bottom left to the top right:
F = k q^2/ r^2 = 1.8 x 10^6 N
Therefore, the top right charge is repelling the bottom right charge by 1.8 x 10^6 N at a 45 degree angle south of west.
3) I added up the vectors to find a resultant vector of the force acting on the bottom left charge: (tr = top right)
a) F(tr)x = 1.8 x 10^6 (cos 45) = 1.27 x 10^6 N (in the negative x direction)
b) F(tr)y = 1.8 x 10^6 (sin 45) = 1.27 x 10^6 N (in the negative y direction)
The resultant vector consists of:
F(x) = (1.27 x 10^6) + (3.6 x 10^8) = 3.61 x 10^8 (negative x direction)
F(y) = 3.61 x 10^8 (negative y direction)
Resultant force = sqrt [(3.61 x 10^8)^2 + (3.61 x 10^8)^2]
= 5.1 x 10^8 N (at 45 degrees south of west)
I'm not sure where to go from here.
I'm wondering if I should have calculated work from each of the separate charges acting on the arbitrarily picked bottom left charge and added them up?:
a) W = Fd = (3.6 x 10^8 N) x (0.025 m) = 9.0 x 10^6 J
+
b) W = Fd = (3.6 x 10^8 N) x (0.025 m) = 9.0 x 10^6 J
+
c) W= Fd = (1.8 x 10^6 N) x (0.035m) = 6.3 x 10^4 J
Total Energy = 1.81 x 10^7 J
Then the TOTAL ENERGY for all 4 charges should be 1.81 x 10^7 multiplied by 4? (Because there are 4 charges):
= 7.23 x 10^7 J[/B]
HELP PLEASE! TEST ON MONDAY!