Why Does the Electric Field Sum Instead of Cancel with Opposite Charges?

[hello721]

Homework Statement

If there are two charges positive and negative and their electric field point in the same direction then the total electric field would be their sum of magnitudes. Why don't we consider the sign of the charges? For example, a parallel plate capacitor is inside the region where both the positive and negative are in the same direction but the Enet is the sum of both of them rather than canceling it out. I was assuming since one charge is negative and one is positive we can just cancel them out since they are equal in magnitude.

Relevant Equations

N/A

If there are two charges positive and negative and their electric field point in the same direction then the total electric field would be their sum of magnitudes. Why don't we consider the sign of the charges? For example, a parallel plate capacitor is inside the region where both the positive and negative are in the same direction but the Enet is the sum of both of them rather than canceling it out. I was assuming since one charge is negative and one is positive we can just cancel them out since they are equal in magnitude.

 

[tech99]

We find the strength and direction of an electric field by measuring the force on a positive test charge. If it is placed between the plates of a capacitor, the negative plate attracts it and the positive plate repels it, so they act in unison. By contrast, if we place the charge outside the capacitor, and some some distance away, the positive and negative plates now push and pull, so acting against each other. For this reason the field is weak outside the capacitor.

 

[topsquark]

Homework Statement:: If there are two charges positive and negative and their electric field point in the same direction then the total electric field would be their sum of magnitudes. Why don't we consider the sign of the charges? For example, a parallel plate capacitor is inside the region where both the positive and negative are in the same direction but the Enet is the sum of both of them rather than canceling it out. I was assuming since one charge is negative and one is positive we can just cancel them out since they are equal in magnitude.
Relevant Equations:: N/A

If there are two charges positive and negative and their electric field point in the same direction then the total electric field would be their sum of magnitudes. Why don't we consider the sign of the charges? For example, a parallel plate capacitor is inside the region where both the positive and negative are in the same direction but the Enet is the sum of both of them rather than canceling it out. I was assuming since one charge is negative and one is positive we can just cancel them out since they are equal in magnitude.

I we add two vectors that act in the same direction we simply add the magnitudes. The signs of the charges are already taken into account here: If the charges had the same sign then the vectors would point in the opposite direction and we'd have to subtract. That's all this problem is doing.

-Dan

 

[Redbelly98]

You've gotten some good replies already. I'll just add: Don't just assign "+" or "-" signs to the electric field value based on the charge(s) producing it. Instead, think about the direction that the field vector (arrow) is pointing -- either away from or toward the charge that is responsible for it, based on the charge's sign.

 

[kuruman]

To write the electric field in a way that its direction will be sorted out automatically for an arbitrary charge, you have to use the formal expression for the electric field. Let's start with a charge on the x-axis at the origin. The formal expression for the electric field on the axis is $$E=\frac{k~q~x}{|x|^3}.$$Now this may seem a silly thing to write, but bear with me and let's see how it works.
Case I: Positive charge $q=|q|$
$$E=\frac{k~|q|~x}{|x|^3}.$$When the point of interest is to the right of the charge ($x>0$), this yields $E=\dfrac{k~|q|~(x)}{|x|^3}=\dfrac{k~|q|}{x^2}.$
When the point of interest is to the left of the charge ($x<0$), this yields $E=\dfrac{k~|q|~(-x)}{|x|^3}=-\dfrac{k~|q|}{x^2}.$
So the field points away from the positive charge as it should.

Case II: negative charge $q=-|q|$
$$E=\frac{k~(-|q|)~x}{|x|^3}.$$When the point of interest is to the right of the charge ($x>0$), this yields $E=\dfrac{k~(-|q|)~(x)}{|x|^3}=-\dfrac{k~|q|}{x^2}.$
When the point of interest is to the left of the charge ($x<0$), this yields $E=\dfrac{k~(-|q|)~(-x)}{|x|^3}=+\dfrac{k~|q|}{x^2}.$
So the field points towards the negative charge as it should.

Of course what is true along the x-axis, is also true along the y and z axes. Superposition then allows us to write the electric field vector at position vector $\mathbf{r}$ due to a charge at the origin as $$\mathbf{E}=\frac{k~q~\mathbf{r}}{|\mathbf{r}|^3}$$ where $q$ can be positive or negative.

To complete the picture, when the source charge is not at the origin but at position vector $\mathbf{r}'$ relative to the origin, we have the most general expression for the electric field from a point charge, $$\mathbf{E}=\frac{k~q~(\mathbf{r-r'})}{|\mathbf{r-r'}|^3}.$$

 

[Redbelly98]

Case I: Positive charge $q=|q|$
.
.
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When the point of interest is to the left of the charge ($x<0$), this yields $E=\dfrac{k~|q|~(-x)}{x^3}=-\dfrac{k~|q|}{x^2}.$
So the field points away from the positive charge as it should.

Should there be absolute-value bars around the "x" for the "x^3" term in the denominator? (And similarly for Case II, to the left of a negative charge.)

 

[kuruman]

Should there be absolute-value bars around the "x" for the "x^3" term in the denominator? (And similarly for Case II, to the left of a negative charge.)

Good catch! I edited the post. Thanks.