Two positive point charges q are located on the y-axis at y = ...

Question

Two positive point charges q are located on the y-axis at y = ±a. Symmetry dictates that the electric field along the x-axis has only an x-component: E_y=E_z=0. Find an expression for E_x if x≪a.

Accepted Answer

Hey, everyone, let's go through this practice problem. Three point charges are located on the X axis at X equals negative A zero. And A, I'll actually draw a picture of this as we go along. So this is the zero point, this is A and this is negative A, the charge at each, the charge at the origin is negative and has a magnitude of Q here. While the other two charges are positive and have a magnitude of two Q each find an expression for the electric field at a point P located on the Y axis such that Y is much smaller than A, we have a two Q here and there is some point P on the Y axis. And we're looking for an expression for the electric field at point P. So the fact the problem specifically tells us such that Y is much smaller than a kind of tells us that we're looking for more of an approximation here rather than exact, rather than an exact value. So the first thing to recognize is that whatever method we use, we should not be afraid to use approximations. But yeah, so this height from the origin to point P is going to be our Y distance. And this must be much greater than a and whatever our final answer is as for our method, an easy way to find the electric field is to recall that the electric field is equal to the negative gradient of electric potential. So something we can do here is find a formula for the electric potential at point P and then take the negative gradient of that potential to find an expression for the electric field. So let's also recall that electric potential is equal to the Coolum constant multiplied by the charge setting up the potential divided by R the distance between the point creating the potential and the point we're looking at or in our purpose for our purposes, R represents the distance between point P and each of the charges. So the total potential at point P is just going to be a sum of the individual potentials set up by the other three particles. So going from left to right here, the potential from the left particle, the leftmost particle at a position of negative A is going to be equal to two multiplied by K multiply the charge because of the fact that the charge is two Q and this is divided by the distance between particle P and that charge. So we're dividing by this distance from P to negative A, we can represent that distance using the Pythagorean theorem. Since one of the legs is Y and another one of the legs has a length of A. So this means that the distance between particle P between point P and that charge is the square root of A squared plus Y squared. So that's where we're putting the denominator the square root of A squared plus Y squared. Then we add the potential from the rightmost charge which is gonna be the exact same magnitude because of the symmetry of the situation. Two KQ divided by the square root of A squared plus Y squared. And then lastly, we then add the potential from the middle particle at the origin which has a charge of a negative Q. So we're going to subtract KQ divided by Y which is the distance between point P and the origin. Now we're going to want to start simplifying this a little bit. Since we have two terms that have the KQ divided by the square root of A squared plus Y squared, we can combine those two terms into one with a four KQ in the numerator. And also to get that square root out of the denominator, we can bring that up into the numerator by inverting the exponent. So we can rewrite that as four KQ multiplied by A squared plus Y squared, all raised the power of negative one half. And then we subtract KQ divided by Y as usual. Now, technically, physically, we already could try if we wanted to, to take the gradient of this and then use that to find the electric field formula. But if you look at our formula, you might notice that this is a pretty tough formula to work with. And it's gonna be a hard one to simplify. So I wanna try and simplify this further before we start taking derivatives. And we can do this by taking advantage of the fact that we're looking for the case where Y is much smaller than A, we can, we can get this into a nicer form without an exponent and without a square root by using a binomial expansion. And binomial expansions work when we're trying to approximate the lead, the main variable as being really small. So we're allowed to use a binomial expansion here and a result will be valid. So the first thing we need to do is turn this A squared into a one. And we can do that by factoring the A squared out of this little parenthetical statement. So we can rewrite this and I'm moving over to the left here. So we can rewrite that part as a squared multiplied by. And then in another set of parentheses, one plus Y squared divided by A squared. And all of that is raised the power of a negative one half and then minus KQY and then raising the A squared to the power of negative one half, we can get this to look a little bit neater by having it written as four KQ divided by A, all multiplied by one plus Y squared, divided by A squared, raise the power of negative one half and then minus a qy. And now we're ready to apply the binomial expansion to simplify this. I recall that for most binomial expansions, we really don't need to expand past the second term. So all we need to do really is get rid of the exponent. But then multiply it by the second term in the sequence. So that's one plus Y squared divided by A squared multiplied by negative one half. So this becomes a minus Y squared divided by two A squared and then this is minus KQ divided by Y. So now we have our formula for the electric potential about as simplified as it's gonna get. So let's get to work with taking the gradient, the negative gradient define the electric field. And specifically, we're only going to worry about the Y component since point P is located just above the origin, it, it only has a displacement on the Y axis. This is the negative gradient of V using Y as our variable. And recall that to take the gradient, we just take the partial derivative of the, of the equation with respect to the variable we're vectorizing. So with a, with a wide hat unit vector is relevant because whatever our result is going to be, it's going to be in the Y direction. But in our case, we are taking the partial derivative of the entire thing, we just uncovered four KQ divided by a multiplied by one minus Y squared, divided by two A squared, all subtracting KQ divided by Y. This is this is a pretty complicated derivative right now. So let's simplify this down a bit by taking the derivative of each of these two components individually. So for the first component, let's pull out the four KQ divided by A. So this becomes negative four KQ divided by a multiplied by the partial derivative with respect to Y of just one minus Y squared divided by two A squared. Then we subtract the negative derivative with respect to why in the Y direction of KQ divided by Y and right off the bat, there are a few things we can simplify here. These negative signs cancel out when the whole thing just becomes a positive. And now it's plain to see that these derivatives actually won't be too hard to evaluate. Now, let's do it. We've got the negative four KQ divided by A and then the derivative in this first derivative right here, the derivative of one is just zero and then the derivative of negative Y squared divided by two A squared. So this is just a basic power rule thing. So we take the exponent from the Y squared and remove that and then multiply the entire term by two, that's a negative two Y divided by two A squared. And we can see these twos will just cancel out and also these negatives will just cancel out. So that's really nice. And then plus the partial derivative with respect to Y of KQ divided by Y and the Y here is in the denominator. So this term is actually going to become a negative. So this first component then minus because when you take the power rule with something that is in the denominator, then we raise that exponent in the denominator but make the whole thing negative that's minus A Q divided by Y squared. And that is the same unifor and that's really it all that's left for us is simplifying. So right now this is equal to four KQY then divided by a cubed because this A and this A squared will multiply one another. So divided by a cubed and then minus KQ divided by Y squared. All of this with a Why hat Unive vector? And we can simplify this even further by getting one single denominator using the, the least common denominators method. So I'm going to multiply the first term, both the numerator and the denominator by Y squared. So that leaves us with four KQ Y cube divided by a cubed multiplied by Y squared and doing the exact same thing to the second term multiplying both the numerator and the denominator by a cubed. Then we end up with adding to the numerator, A negative KQ A cubed and that is our entire formula all in the Y direction. So we add the cute little white hat here. And so that's the answer to the problem. This is our formula or the electric field at point P. It's a pretty tough problem and it involves a binomial expansion. So that's probably gonna throw some people off. So don't feel too bad if you couldn't figure that part out. But yeah, but this is a pretty tough problem and that's how you would solve it. And that's all for this video. I hope it helped you out. And if it did, please consider checking out some of our other videos which will give you more experience with these types of problems, but that's all for now. And I hope you all have a lovely day. Bye bye.

Two positive point charges q are located on the y-axis at y = ±a. Symmetry dictates that the electric field along the x-axis has only an x-component: E_y=E_z=0. Find an expression for E_x if x≪a.

Key Concepts

Electric Field

Superposition Principle

Symmetry in Electric Fields

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Two positive point charges q are located on the y-axis at y = ±a. Symmetry dictates that the electric field along the x-axis has only an x-component: Ey=Ez=0. Find an expression for Ex if x≪a.

Key Concepts

Electric Field

Superposition Principle

Symmetry in Electric Fields

Watch next

Two positive point charges q are located on the y-axis at y = ±a. Symmetry dictates that the electric field along the x-axis has only an x-component: E_y=E_z=0. Find an expression for E_x if x≪a.