Question

This following question has been bugging me for a while as I Do not know how exactly can I find the value

Two charges, q1 and q2, 3nC and -3nC respectively are a distance of 10 cm from each other. Find the magnitude of the electric field at point which sits 5 cm above the midpoint between q1 and q2.

I have figured out that by using Pythagorean theorem I have got the distance between the points and p. Since q1 and q2 are polar opposites, and at equal distances from point p.
I have tried to find the field for atleast 1 of the points. See more details in replies.

Answer 1

I tried to map out the Electric Field of one points but I swear I'm doing something completely wrong.

So far, I've used the to find the Electric field of point p in respect to q1,
k being Coloumb's constant rounded up.
\[E _{q_{1}|p} = \frac{ 1 }{ 4 \pi k } \frac { q_{1} }{r}\]
But Im' getting such a small number that I don't believe this to be the right formula.
\[3.75 * 10 ^{-21}\] for the electric field of a single point

While the answer to this question is according to the paper, is 7.6 kV/m s, which is, definetly quite big in comparison to this point.

Answer 2

I refined and got a more practical equation, which simply is \[E = \frac{kQ}{d^{2}} \] this seems to get much better results, as well as correcting some Exponential rule (forexample I was already applying exponents within my pythogorean theorem

But the resulting result is 3.81 N/C. Im not quite sure but this can this be converted to V/m?
since if this is true, multiplying this by 2 (as there are two fields) and by their magnitude it would be aprx 7.6 V/m.

However this still would 10^3 units too short. Anyone know why this is?

Answer 3

N/C and V/m are just two ways to name the same unit.

Answer 4

That however does not explain why the answer is 7.6 kV/m and not 7.6 V/m, which I got. Notice the k in front of the V in the answer.

Answer 5

Orders of magnitude prove that kV is the right answer.
for 3 nanocoulombs at distance 5 cm, you get:

\(9.10^9 \times 3.10^{-9} / 0.05^2 = 11 kV/m\)

Answer 6

I have a feeling this is just giving me a major headache, You are definitely right about the order of magnitudes and that's a mistake I did do clearly on my calculations, but the direction was incorrect as 5 cm is not the line between p and a charge

To redraw,


So far this is my re-work:
using pythogorean theorem
\[d_{3}^{2} = 0.005^{2} + 0.005^{2}\]\[d_{3} = \sqrt{0.05^{2} + 0.05^{2}} = \sqrt{0.0025 + 0.0025} = \sqrt{0.005}\]\[\frac{ 8.99*10^{9} * 3 *10^{-9} }{ \sqrt{0.005} ^{2}} = \frac{ 26.97 }{ 0.005 } = 5394 \frac{ V }{ m } = 5.39 \frac{ kV} {m}\]

but what I still do not get, why the paper says the answer is 7.4 kV/m
Source: http://ta.ramk.fi/~jouko.teeriaho/Electricity&Magnetism.pdf

Answer 7

arg, this needs and edit, the diagram above was missing nano's for the Coloumbs

Answer 8

Im just starting to think there is a mistake in the paper it self. Just going to ignore it, as what I got is pretty much 10.78 kV/m, which when rounded up is the same as Vincent's Answer. But if someone can within a few hours say why the answer it self is written as 7.4kV/m on the paper, and correct my mistake, it will be appreciated and theyll get a medal.

Answer 9

The correct answer is 7.6 kV/m
My previous one was just an estimation of the order of magnitude.

Answer 10

Your 5.39 kV/m is correct in that it is the field created by one of the charges only.

As there are 2 charges, exactly opposite one another, acting at right angles at point P, their electric fields add up vectorially, and the final result is your 5.39 multiplied by \(\sqrt 2\)