# Electric field due to two positive charges

• sebah
In summary: That's right. The Electric field at point P is just the vector sum of the fields from the two charges.
sebah
I am having trouble solving the following problem. I am given two positive charges on the x axis:

I know that the electric field strength at point P is ##E=150 \frac{V}{m}##, ##d=1.8m## and ##a=2.5m##. I want to find the charge of ##Q##. As far as I know, the electric field on the y-axis between the two charges is given by:

$$E=\frac{2xQ}{4 \pi \epsilon_0 (x^2+y^2)^{\frac{3}{2}}}$$

where ##x## is the distance between one charge and the origin and ##y## is the distance from the origin on the ##y##-axis. In this case ##y=d## and ##x=\frac{a}{2}##.

$$E=\frac{aQ}{4 \pi \epsilon_0 ((\frac{a}{2})^2+d^2)^{\frac{3}{2}}}$$

Solving for ##Q##:

$$Q=\frac{E 4 \pi \epsilon_0 ((\frac{a}{2})^2+d^2)^{\frac{3}{2}}}{a}$$

Plugging in values gives me:

$$Q=7.02 \cdot 10^{-8} C=70nC$$

The correct answer is supposed to be ##~50nC=50\cdot 10^{-9}C##. Am I doing something wrong or is there a mistake in the solution?

Does your value for E make sense in the limit x=0??

hutchphd said:
Does your value for E make sense in the limit x=0??

Not really I think. If x=0 which means that the distance between the charges goes to zero (a=0) I get E=0. Not sure why I am getting zero here though. Should be something like the electric field of 2Q at the origin.

If I were doing it, I would just do the vector sum of field from each charge. What is the net vector at point P?

sebah said:
Not really I think. If x=0 which means that the distance between the charges goes to zero (a=0) I get E=0. Not sure why I am getting zero here though. Should be something like the electric field of 2Q at the origin.
Where did you get the formula? Its wrong.

scottdave said:
If I were doing it, I would just do the vector sum of field from each charge. What is the net vector at point P?

That's exactly what I did. My formula looks just like the one derived by R. Shankar in his yale physics course. LINK

hutchphd said:
Where did you get the formula? Its wrong.

Like I mentioned in the other reply I got the formula from R. Shankar and his yale course. LINK

In the lecture, he is calculating the field of a dipole where one charge is positive and one is negative. The y-components cancel out and the x-components add.

That's not the case here. Draw the vectors representing the fields from those two identical positive charges. They do not cancel out in the same way as the fields in your link.

Moral: Don't take formulas blindly from one situation and apply them to a totally different situation.

Last edited:
sebah
RPinPA said:
In the lecture, he is calculating the field of a dipole where one charge is positive and one is negative. The y-components cancel out and the x-components add.

That's not the case here. Draw the vectors representing the fields from those two identical positive charges. They do not cancel out in the same way as the fields in your link.

Moral: Don't take formulas blindly from one situation and apply them to a totally different situation.

Oh I see. In my case the horizontal components cancel and the vertical components add. I will then get:

$$E=\frac{2*y*Q}{4 \pi \epsilon_0 (x^2+y^2)^{\frac{3}{2}}}$$

Hint: an electric field is a vector!

Incidentally, you should be able to write down the electric field from point charges from first principles. In fact the easy way to do this is to notice the symmetry, write down the electric (scalar) potential and take the derivative d/dy to get the field in that direction. But any correct way is OK.

## 1. What is the formula for calculating the electric field due to two positive charges?

The formula for calculating the electric field due to two positive charges is given by E = (k*q1*q2)/r^2, where k is the Coulomb's constant, q1 and q2 are the magnitudes of the two charges, and r is the distance between them.

## 2. How does the magnitude of the electric field change as the distance between the two positive charges changes?

The magnitude of the electric field decreases as the distance between the two positive charges increases. This is because the electric field follows an inverse square law, which means that the strength of the field decreases as the distance from the source increases.

## 3. What is the direction of the electric field due to two positive charges?

The direction of the electric field due to two positive charges is away from the charges. This means that the field lines point outward from the positive charges.

## 4. Can the electric field due to two positive charges be zero?

Yes, the electric field due to two positive charges can be zero. This occurs when the two charges have equal magnitudes and are placed at equal distances from a point in between them. In this case, the electric fields from the two charges cancel each other out, resulting in a net electric field of zero.

## 5. How does the electric field due to two positive charges affect a third charge placed in the vicinity?

The electric field due to two positive charges exerts a force on any other charges placed in the vicinity. This force is attractive if the third charge is negative and repulsive if it is positive. The strength of the force depends on the magnitude of the third charge and its distance from the two positive charges.

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