When the voltage across C is 5.6V the switch is quickly thrown to position B. At

Question

When the voltage across C is 5.6V the switch is quickly thrown to position B. At what time after the switch is thrown to B will the voltage across C be equal to IV? Three point charges q_1, q_2, and q_3 are fixed at the vertices of an equilateral triangle whose sides are 2.50 m-long. Find the potential due the system of these charges (q_1= q_2= q_3 = +4.20 mu C) at point P, midway between q_2&q;_3 If a forth particle that has a charge +e is placed at point P and released from rest, find its speed when it is a very far from the origin.

Explanation / Answer

potential due to a point charge at distace r is given by

V = kq/r
distance of q1 fromP = 2.50sin60 = 2.165 m

distance of q1 and q3 = 2.5/2 = 1.25 m
so Vnet = [ 9x 10^9 x 4.20 x 10^-6 / 2.165] + [ 9x 10^9 x 4.20 x 10^-6 / 1.25] + [ 9x 10^9 x 4.20 x 10^-6 / 1.25]
= 77939.6 Volt

b) Potential difference when charge is at P and infinity.

deltaV = 77939.6 - 0 = 77939.6 volt

and KE = q deltaV

v = sqrt[2 q deltaV / m ]

mass is not given, I am considering it as proton

them m = 1.67 x 10^-27 kg

putting values,

v = 3.86 x 10^6 m/s

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