Some Electric Potential energy questions

[wr1015]

A charge of 3.19 µC is held fixed at the origin. A second charge of 3.19 µC is released from rest at the position (1.15 m, 0.530 m) .

(a) If the mass of the second charge is 3.10 g, what is its speed when it moves infinitely far from the origin?

ok so I've broken it up into x and y components and used pythagorean theorem to find the line that represents the force that the second charge exerts on the fixed charged $$\sqrt{1.15^2 + .530^2}$$ and my answer was 1.266 m. Then i used tan$$^-1$$ (.530/1.15) to find theta which came out to be 11.98$$^0$$. Then i calculated 1.266 (cos 11.98) for the positive x-direction and 1.266(sin 11.98) for the positive y-direction and i got 1.2384 m for the x and .26278 m for the y. Then i plugged those values back into pythagorean theorem again to find the net force $$\sqrt{.26278^2 + 1.2384^2}$$ and got 1.2666N. Then i used the Force per unit charge to find the net electric field and got (1.2666/3.19E-6) = 3.97E5 and multiplied by 2 since there are 2 charges and got 7.94E5 as my net electric field. And then i plugged that value into $$\sqrt{2q\Delta V/m}$$ to find velocity. Obviously I'm doing something wrong somewhere because I'm not getting the right answer... any help would be greatly appreciated.

 

[marlon]

What's the right answer ?

Hint : what is the potential if the charge is that far away from the first charge.

marlon

 

[wr1015]

What's the right answer ?

Hint : what is the potential if the charge is that far away from the first charge.

marlon

well that's what I'm trying to figure out..

 

[marlon]

Well,

what's the forumula for the potential ?

Again, what's the right answer ? You told in your first post that you keep getting the wrong answer ? How do you know this ?

marlon

 

[wr1015]

Well,

what's the forumula for the potential ?

Again, what's the right answer ? You told in your first post that you keep getting the wrong answer ? How do you know this ?

marlon

formula for potential is $$\Delta U/q$$ but don't you have to figure out net electric field first since $$\Delta U = qEd$$ ??

i'm using webassign so the answers i keep entering are wrong

 

[marlon]

formula for potential is $$\Delta U/q$$ but don't you have to figure out net electric field first since $$\Delta U = qEd$$ ??

i'm using webassign so the answers i keep entering are wrong

The formula for the potential of a charge q felt at a distance r is $$U = k \frac{1}{r}$$ where k is a constant that does not matter here. Now, what's the U if q is at spatial infinity ?

It is important to understand the concept of potential (which is the aim of your exercise). A potential is actually potential energy per unit of electric charge. So two electric charges q1 and q2 have potential energy $$k \frac{q1*q2}{r}$$ where r is the distance between those two charges. If you divide this expression by, let's say, q1 you get the potential of q2. This potential, actually is a measure of how big the interaction (ie the magnitude of the potential energy) with an electrical charge will be if you put some charge q1 on a distance r from q2.



marlon

 

[wr1015]

The formula for the potential of a charge q felt at a distance r is $$U = k \frac{1}{r}$$ where k is a constant that does not matter here. Now, what's the U if q is at spatial infinity ?

It is important to understand the concept of potential (which is the aim of your exercise). A potential is actually potential energy per unit of electric charge. So two electric charges q1 and q2 have potential energy $$k \frac{q1*q2}{r}$$ where r is the distance between those two charges. If you divide this expression by, let's say, q1 you get the potential of q2. This potential, actually is a measure of how big the interaction (ie the magnitude of the potential energy) with an electrical charge will be if you put some charge q1 on a distance r from q2.



marlon

thank you for your help, i understand now..