Question

Guys I'm in kind of a PICKLE!!!!!! I know people say it a lot but I will give Brainiest to the best explained answer. Determine the net force charge acting at q1 (+ 2.0 × 10^-5C), caused by q2 (-4.0 × 10-5 C) and q3 (-4.0 × 10^-5C). They create a right angles triangle, where q1 is at the 90° corner
Determine the net electric field acting at q1

Answer 1

Answer:

E≅1.2×10^7 N/C

Explanation:

First off I'd like to say that I'm taking "net electric field" to mean that they don't want this answer to be put into vector component form and instead want magnitudes. Sometimes the wording of these questions throws me off, so sorry ahead of time if that's what they want from you!

Edit: I ended up adding it anyways ;P

Since we are observing the net electric field acting at q1, we need to use the formula:  $E=k\frac{q}{r^{2} }$

And since we are observing the effects of multiple charges at once...

E=ΣE, which just means wee need to add all the observed electric fields together:

ΣE= $k\frac{q2}{r^{2} } +k\frac{q3}{r^{2} }$

Since we are observing [static] electric fields here, we don't actually need q1's charge. (Though if you wanted to find the net force you would.) Now, before we start plugging values in, let's acknowledge what we know. We know that:

• q2=q3
• they are the same distance from q1

These are actually really nice to have, because now we can simplify our expression to:

$E=k\frac{2q}{r^{2} }$

Now let's plug in our values and get an answer out.

E= 2(8.99×10^9)(4×10^-5)/(0.24)

Plugging all that in, I get:

E≅1.2×10^7 N/C

If you end up needing the net force, F=(q1)(E). That is, you just multiply the electric field by the value of q1. And again, if your teacher wants the answer in vector component form, then the answer will look different.

Let me know what doesn't make sense, or if I got something wrong. Good luck with AP Phy.!

Edit: I put the component form for my answer in the attachment. I also noticed a small calculator related error in my original answer. I updated that to match the new one.