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no idea where to start, but i want to learn so steps would really help. thank you?
So identical point charges (q = +7.40 10-6 C) are fixed at diagonally opposite corners of a square with sides of length 0.465 m. A test charge (q0 =-2.30 10-8 C), with a mass of 6.65 10-8 kg, is released from rest at one of the empty corners of the square. Determine the speed of the test charge when it reaches the center of the square
1 Answer
- ?Lv 76 years ago
The two diagonally opposite charges (each charge q) will cause the potential at one of the empty corners to be
V = k q / a + k q / a = 2 k q / a
where k = 1/(4 pi eps0) and a is the length of a side of the square.
The potential they cause at the center of the square (at a distance a/sqrt(2) from them) is
V = k q / (a/sqrt(2)) + k q / (a/sqrt(2)) = 2 k q /(a/sqrt(2)) = 2 sqrt(2) * k q / a
Therefore the change in potential energy for the test charge Q moving from a corner to the center of the square is
delta(U) = Q * delta(V) = Q * 2 k q /a *( sqrt(2) -1)
Energy conservation requires that this change (loss, because Q<0 and q>0) in potential energy goes into increasing the kinetic energy of the test charge from 0 to 1/2 m v^2, so
delta(K) = - delta(U)
1/2 m v^2 = - Q * 2 k q /a *( sqrt(2) -1)
v = sqrt( -4 k Q q (sqrt(2)-1) / m)
You can fill in the numbers I guess.