- #1
Kudo Shinichi
- 109
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HELP!A question on electricity (electric field and electric potential)
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?
a)E=KQ/r^2
Etotal=(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
Vtotal=(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=KQ1Q2/r
Vtotal=(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
Vtotal=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.
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
Etotal=(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
Vtotal=(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=KQ1Q2/r
Vtotal=(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
Vtotal=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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