Finding the electric potential (voltage) of a point charge? I\'ve figured out pa

Question

Finding the electric potential (voltage) of a point charge?

I've figured out parts a. and b.

How do I do part c? It looks like Bob's equation is the standard V = (kQ)/r equation but with the added constant.

To solve part c., I tried adding Bob's constant to the answer I got in part a (76770 + 61416),. but that doesn't work, and I don't know how to solve this. Help please!

The three charges are held in place in the figure below, where L = 1.25 m +7.45 .C +2.75 C -1.72 C (a) Find the electric potential at point P, at infinity, and the magniude of their difference VP =76770 Voo-10 AVI = 76770 (b) Suppose that a fourth charge, with a charge of 5.89 C and a mass of 4.71-10-3 kg, is released from rest at point P. What is the speed of the fourth charge when it has moved infinitely far away from the other three charges? Vf = 13.85 m/s (c) Suppose your strange friend Bob insists his equation for the electric potential of a point charge is better: where C = Bob's constant, based on Bob's favorite number = 61416 V What would be Bob's answer for the above questions? for parta VP 138186 Va = 61416 for partb Vf= (Which answers changed? Which are important changes? Is Bob's formula better or worse than yours?) m/s

Explanation / Answer

Vp = [ (9 x 10^9 x 2.75 x 10^-6 / 0.625) + 61416 ] + [ (9 x 10^9 x-1.72 x 10^-6 / 0.625) + 61416 ]

+ [ (9 x 10^9 x 7.45 x 10^-6 / 1.25sin60) + 61416 ]

= 261018 Volt

V~ = 3(61416) = 184248 volt

deltaV = Vp - V` = 261018 - 184248 = 76770 volt

deltaV is same hence vf will also be same.

vf = 13.85 m/s

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