three identical point particles that have charge q are at the vertices of an equ
ID: 1692139 • Letter: T
Question
three identical point particles that have charge q are at the vertices of an equilateral triangle that is circumcised by a circle of radius a that lies in the z=0 plane and is centered a the origin. The values of q are a are at 3.00uC and 60.09 cm respectively. (Assume the potential is zero very far from all charges)a. what is the electric potential at the origin?
b. what is the electric potential at the point on the z axis at z=a?
c. How would your answers to parts (a) and (b) change if the charges were still on the circle but one is no longer at a vertex of the triangle? explain your answer.
Explanation / Answer
Given that charge q = 3.00 x 10^-6 C distance r = 0.6009 m --------------------------------------------------------------- a) the electric potential at the origin is V = kq/r + kq/r + kq/r = 3 kq/r = 3(9.0 x 10^9 N.m2/C2)(3.00 x 10^-6 C)/( 0.6009 m) = 1.3479x 10^5 V = 1.3479x 10^2 kV b) The distance from the each charge to the given point is d =v[ r^2 + r^2] = v2 r So the net eletctric potential due to all charges is V =3 kq/d = 3(9.0 x 10^9 N.m2/C2)(3.00 x 10^-6 C)/(v2( 0.6009 m) ) = 9.531644 x 10^4 V = 95.3644 kV c) In this case potential of part (a) and (b) are remains constant.
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