- #1

Kudo Shinichi

- 109

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**HELP!A question on electricity (electric field and electric potential)**

## Homework Statement

Three point charges are located at the corners of an equilateral triangle as shown.

http://s5.tinypic.com/2hdvalh.jpg

a) Find the magnitude and direction of the electric field at P

b) Find the magnitude and direction of the electric field at P(The potential at infinity is zero)

c) What is the electric potential energy of a particle with charge 4 µC placed at P?

d) If the mass of the particle is 7.0*10^-11kg, what minimum velocity muse it be given so that it can escape to infinity?

## The Attempt at a Solution

a)E=KQ/r^2

E

_{total}=(9*10^9x1*10^-6)/0.3^2+(9*10^9x2*10^-6)/0.3^2+(9*10^9x-2.5*10^-6)/0.52^2=216549.06 point up because the top point has the strongest charge (I am not sure whether I got the direction right or not)

b)V=KQ/r

V

_{total}=(9*10^9x1*10^-6)/0.3+(9*10^9x2*10^-6)/0.3+(9*10^9x-2.5*10^-6)/0.52=46678.85

c)V=KQ

_{1}Q

_{2}/r

V

_{total}=(9*10^9x1*10^-6x4*10^-6)/0.3+(9*10^9x2*10^-6x4*10^-6)/0.3+(9*10^9x-2.5*10^-6x4*10^-6)/0.52=0.187

d)V=sqrt(2kQ/mr)

m=mass

V

_{total}=sqrt((9*10^9x1*10^-6)/(7*10^-11x0.3))+sqrt((9*10^9x2*10^-6)/(7*10^-11x0.3))-sqrt((9*10^9x2.5*10^-6)/(7*10^-11x0.3))=35500654.39

For part d I don't really know how to solve for velocity after I got the voltage

Even though I got the answer for all of them, but I don't think it is the right way to approach the problem, because the answers are either too big or too small

Please help me with it. Thank you very much.

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