Question

Please help me out?
12a: Two charges are held in place as shown. Determine the magnitude and direction of
the electric field at the origin. Use Q = 3 nC, q = −2 nC, D = 4 m, L = 2 m and H = 5 m.
b: Determine the magnitude and direction of the acceleration of a 2 µC charge of mass 10−25
kg which is released at the origin.

Answer 1

Ok, so first things first, they want you to find the electric field at the origin. What's the electric field?

\[\vec E = k \frac{q}{r^2} \vec r\]

And \(\vec r\) is purely just a unit vector pointing in the direction of the electric field while the \(k \frac{q}{r^2}\) is the magnitude of the electric field.

So give it your best shot at finding either of these, and I'll try to help you if/when you get stuck.

Answer 2

alright so for Q:
\[E=\frac{8.99*10^{9}Nm^{2}}{C^{2}}*\frac{3nC}{(4m)^{2}}*\frac{10^{-9}C}{1nC} \]
\[=\frac{1.69N}{C}\] does this look right?

Answer 3

Yeah this looks really really good, I like seeing units this is awesome I love being able to read someone's physics thank you haha. So far so good!

So this is the magnitude part, now you just need to find the direction of the electric field.

Answer 4

I forgot the direction
\[\frac{1.69N}{C}x\]

Answer 5

Yeah that's right, I think the next one is trickier to find, I'm gonna go downstairs and make some coffee and breakfast so give me a minute, I'll be back!

Answer 6

so for q?:
\[E=\frac{8.99*10^{9}Nm^{2}}{C^{2}}*\frac{-2nC}{(29m)^{2}}*\frac{10^{-9}C}{1nC}\]
\[=\frac{-0.021N}{C}\]
I have to find the x and y magnitudes here right. This is where I get lost.

Answer 7

? its been awhile

Answer 8

Yeah this is the most confusing part I agree, but since you've found the magnitude the only thing left to find is the direction. The direction vector is just a vector that points in the direction you want with no length of its own.



Notice that if we multiply it by -1 it just flips direction so it's not too important that it points the correct direction or not when you solve for it.

Now notice, it lies exactly along the hypotenuse of that triangle, in other words it's parallel with the position vector of q,

\[\vec R = L \vec x + H \vec y\]

The only difference between this vector and the unit vector is length. So we just divide by the length of R,

\[\vec r = \frac{\vec R}{|\vec R|}= \frac{\vec R }{\sqrt{\vec R \cdot \vec R}} \]

I hope that answers your question of how to find the unit vector, I feel like I was kind of long winded there sorry haha.