Find the potential. I was told to use Kq/R = V The problem # I need help with is

Question

Find the potential.
I was told to use Kq/R = V

The problem # I need help with is #12.
thank you for your help!

, , ) too, to (10) through an elec o thrh an electric feld given by the one in the prevous problem ome oving & charge q along a path given by the function y 3x from tric eld given by the one in the previous problerm 11. Take the gradient of the folowing electric potentiat: V 11. Take the gradient of the following electric potential Show that ih is showth.. . HINT: it will be identical to the one we use for the electric field of a point charge. E useful to use the definition of a unit vector Four equal charges of Lud, sit at the corners of a square of side length 1 cm. A fifth equal charge is brought in from far away and placed at the center of the square. How much work does it take to bring in the fifth charge? Find the potential 0.10 m up along the symmetry axis of a charged ring of radius 0.05 m carrying charge of 4.0uC 12 13. or the sake of future exams, it would be a good idea to know how to solve all of the conti charge distributions we discussed in class including the rod (on and off axis), the arc, the ring a disk.

Explanation / Answer

r = 1/sqrt2 = 0.707 m

q1 = q2 = q3 = q4 = 1uc

initial potential energy of fifth charge = Ui = 0

potential at the center

V = k*q1/r + k*q2/r + kq3/r + k*q4/r

V = 9*10^9*4*10^-6/0.707 = 50919.4 V

final energy of fifth charge = Uf = Q*V = 1*50919.4 J = 50919.4 J

work done = change in PE = 50919.4 J <<<--------answer

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