The Problem Data

Q1      P     Q2
+ -----------  -

$Q_1 = +50.0 \mu C$
$Q_2 = -10.0 \mu C$
$d(Q_1, Q_2) = 40.0 cm\ \ \ d(x,y) = distance\ between\ x\ and\ y$

I don't know for sure, it's not in the problem description, but I guess $P$ lies in the middle between $Q_1$ and $Q_2$.


Calculate the electric field size in $P$.

What I've tried

Nothing really, I just don't know how to start.

Thank you in advance, I'm sure this will be far to easy for you guys!


As I was asked if I knew how to solve it if one of the electric charges would be zero, I hereby want to tell I do.

I know:
$F = k \cdot \frac{|Q_1] \cdot |Q_2]}{r^2}$
$E = \frac{F}{|Q_t|}\ \ \ \ (Q_t = test\ charge)$
$= k \cdot \frac{|Q_s| \cdot |Q_t|}{r^2|Q_t|}\ \ \ \ (Q_s = source\ charge; Q_t = test\ charge)$
$= k \cdot \frac{|Q_s|}{r^2}$

Note: test charge and source charge are a free translation of 'proeflading' and 'bronlading' (Dutch/Flemish).

  • 1
    $\begingroup$ First hint: Would you know how to solve the problem if one of the charges $Q_1$ and $Q_2$ happens to be zero? $\endgroup$
    – Qmechanic
    Dec 12, 2011 at 18:16
  • 1
    $\begingroup$ Yes I do, it would be $E = k * \frac{Q}{r^2}$ with $r$ as distance and $k = 8.99 * 10^9 N * m^2/C^2$ $\endgroup$
    – user6617
    Dec 12, 2011 at 18:21
  • 1
    $\begingroup$ You are almost there! Second hint: The total $E$-field is just the 'sum' of the two partial $E$-fields you just calculated. Now recall that $E$-fields (in one dimension) comes with a sign, which denotes the direction of the $E$-field. The signs are the tricky part! $\endgroup$
    – Qmechanic
    Dec 12, 2011 at 18:36
  • 1
    $\begingroup$ Hi lef2, and welcome to Physics Stack Exchange! Unfortunately I do have to tell you that, this question is not appropriate for this site. This is a place for conceptual questions, not "do-my-homework" problems, and just posting your problem and saying "I don't know how to start" falls under the latter category. When you are posting a homework question you need to narrow it down to focus on the specific concept that is giving you trouble; for example, you could have said "I know how to calculate the electric field from one source charge but I am confused by having two" (or something like that). $\endgroup$
    – David Z
    Dec 12, 2011 at 20:33
  • 1
    $\begingroup$ (cont.) Since this question has already been asked and answered and edited to contain some general information, I won't close it, but please keep the guidelines in mind if you post homework questions in the future. You can check our [homework FAQ](meta.physics.stackexchange.com/questions/714) for more information on posting appropriate homework questions here. $\endgroup$
    – David Z
    Dec 12, 2011 at 20:35

2 Answers 2


Keeping this general, I think the important principle is superposition which leads to my prescription below:

Electric fields are caused by charges, at a point the total electric field is equal to the sum of the fields caused by each charge being considered (superposition). Therefore to answer any general question like this I would recommend, counting your charges, working out the individual contributions seperately, then summing them.

Be carfeul with signs!


Let's see Two Charges $q_1$ and $q_2$, We have to find electric field at some point between them.

Let's assume $d$ to be the distance between the charges, which is constant and $x$ (from the center of the line joining the two charges, $\frac{d}{2}$ from the chagres) to be the distance where we want to measure electric field at.

                           |      x
       q_1                 |                 q_2 

Now, $\large E_{q_1x} = \frac{1}{4 \pi \epsilon} \frac{q_1}{\left(\frac{d}{2}+x\right)^2}$ and $\large E_{q_2x} = \frac{1}{4 \pi\epsilon} \frac{q_2}{\left(\frac{d}{2}-x\right)^2}$

Total Electric Field at $x$,

$$E_{x} = \frac{1}{4 \pi \epsilon} \frac{q_1}{\left(\frac{d}{2}+x\right)^2} + \frac{1}{4 \pi\epsilon} \frac{q_2}{\left(\frac{d}{2}-x\right)^2}$$

I tried to make it as generalized as possible so you don't have to face problems with such type of problems in future. I hope it helps you and Good Luck for your Exam!

I tried to attach a picture but because of low reputation I couldn't, Sorry.

  • $\begingroup$ Thank you, I suppose $\frac{1}{4\pi\epsilon}$ is the constant, which is in normal air $8.99 \cdot 10^9$? $\endgroup$
    – user6617
    Dec 12, 2011 at 18:46
  • $\begingroup$ Yeah, you are right! $\endgroup$ Dec 12, 2011 at 18:50
  • 1
    $\begingroup$ Hi Ishann - I've mentioned to you before that giving away complete answers is against our homework policy. In many cases, giving a formula into which the poster can just plug numbers from the problem is pretty close to being a complete answer, even if you do show the derivation. I understand that you're trying to be helpful, and we appreciate the effort, but keep in mind that students are helped more by having to work through the reasoning themselves than by just seeing it done. $\endgroup$
    – David Z
    Dec 12, 2011 at 20:40
  • 1
    $\begingroup$ You're not in trouble or anything, just keep this in mind when you answer homework questions here in the future. Thanks! $\endgroup$
    – David Z
    Dec 12, 2011 at 20:41
  • $\begingroup$ Oh Sorry, It won't happen again; I will make sure that I answer homework questions according to the site's policy. $\endgroup$ Dec 13, 2011 at 3:37

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