(II) A positive charge q is placed at the center of a circular...

Question

(II) A positive charge q is placed at the center of a circular ring of radius R. The ring carries a uniformly distributed negative charge of total magnitude ― Q.

(a) If the charge q is displaced from the center a small distance x as shown in Fig. 21–71, show that it will undergo simple harmonic motion when released.

(b) If its mass is m, what is its period?

Accepted Answer

Hi everyone. Let's take a look at this practice problem dealing with electric fields. So in this problem, a uniformly charged ring with a total charge of minus capital Q is shown in the figure below. It is centered on the origin with a radius R. A positive charge of little Q is placed at the center. If little Q is displaced by a small distance Y along the y axis perpendicular to the ring, we need to determine two different quantities. For part one, we need to find the expression for the spring constant. That would result if the charge undergoes simple harmonic motion when released. And for part two, we then need to determine the period of this motion. If mass of little Q is M below the question, we're given a diagram of what was described in the problem. We're also given four possible choices as our answers. For choice. A for part one, we have low Q multiplied by capital Q divided by the quantity of four pi epsilon knot R squared. And for part two, we have two pi multiplied by the square root of the quantity of pi epsilon knot Mr squared divided by the quantity of lowercase Q multiplied by capital Q for choice B for part one, we have little cube multiplied by capital Q divided by the quantity of four pi epsilon knot R cubed. And for part two, we have four pi multiplying the square root of the quantity of pi epsilon knot. Mr cubed divided by the quantity of little Q multiplied by capital Q for choice C for part one, we have low Q multiplied by capital Q divided by the quantity for pi epsilon knot R cubed. And part two, we have two multiplying the square root of the quantity of pi epsilon knot. Mr cubed divided by the quantity of low Q multiplied by capital Q. And finally, for choice D for part one, we have low Q multiplied by capital Q divided by the quantity of four pi epsilon knot R squared. And for part two, we have two pi multiplied by the square root of the quantity of epsilon not Mr cubed divided by the quantity of low Q multiplied by capital Q. Now, for part one, we need to find the spring constant for the simple harmonic motion that this charge would undergo when released. Now, in order to find the spring constant, we need to find the force acting on this charge. And in order to find the force, we need to find the electric field. So that's where we're going to start. So we call your formula for the electric field uh along a line through the center of a ring due to a ring of charge. And since this is gonna be along in the Y direction, we're just gonna be looking at the Y component of the electric field, we'll have ey is equal to here, we'll have our charge which is minus capital Q multiplied by Y divided by the quantity A four P. Absolutely not multiplying the quantity of Y squared plus R squared in quantity raised to the three halves. Now we're told in the question that we're gonna be looking for a small distance for why? So that means that Y is gonna be much smaller than R. So we can actually simplify this expression when Y is much less than R, we'll actually get ey is equal to minus capital qy divided by or pie. Absolutely not. And inside the parentheses for next term, I'll have the Y squared plus R squared. But since Y squared is really small compared to R squared, it's essentially not going to contribute anything. So I'm just gonna have R squared raised to the three halves which when I do that just gives me R cubed. So now that I have an expression for my electric field, I can now calculate the force. So recall your formula or uh the force in relationship to the electric field that is F, which is our force, it's going to be equal to. And here we're gonna have the charge inside the electric field, this is gonna be the little queue and that's gonna be multiplied by the electric field that it's in, which is gonna be our ey expression that we just found. So we'll plug in for ey, so we'll have F it's equal to minus little Q multiplied by capital Q divided by or pi epsilon knot are cute. And that fractions could be multiplied by Y. Now, we need to determine what the spring constant is. And so for this, we need to recall Hook's Law to recall that Hook's Law is FS equal to minus KX. So here we just need to identify what each of these variables actually represents. So the forces are the same in both expressions or Y is the same as our displacement X in Hook's Law, we have a minus sign out in front in both cases. And so what's left here in this expression is gonna be our K value. And so that means our K, our spring constant is going to be equal to little Q multiplied by capital Q divided by the quantity or pi epsilon, not R CED. So that's gonna be our answer to part one. Now, for part two, we need to find the period. And for that, we're going to recall our formula for the period of oscillation or a mass on the spring to recall that formula that is T is equal to two pi multiplied by the square root of M divided by K for T, here is our period M is gonna be the mass and K is our spring constant. Now, we just found an expression for the spring constant K. So we'll plug that into a formula for the period. We'll have T is equal to two pi multiplied by the square root of M divided by the quantity of little Q multiplied by capital Q divided by or pie absent. Not are cute. So we can simplify the fraction inside the square root. We have T is equal to two pi multiplied by the square root of or pie epsilon knot. Mr cubed divided by little Q multiplied by capital Q. And I can simplify this just a little bit more. So I have a four inside the square root so I can move it outside. And so I'll take the square root of four which is two. And so I'll just multiply the factor out of front by two. And so I'll have our period T is equal to four pi multiplied by the square root of pi epsilon N Mr cubed divided by the quantity of low Q multiplied by capital Q. So this is gonna be my answer for part two. And so now I have an answer for both parts. I can look at my answer choices and this actually corresponds to answer choice B. So just to recap real quick what we've done, we start off for part one, calculating the electric field due to the ring of charge. And then we use that lecture field along with our other point chart, the little Q to calculate the force. And then related that expression to that of Hook's Law in order to determine what our constant was. Once we have our spring constant. For part two, we just use our formula for the period of oscillation for a mass on a spring. And in place of the spring constant, we plugged in our formula that we determined from part one and then just simplified the expression. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Coulomb's Law

Simple Harmonic Motion (SHM)

Period of Oscillation

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