Use the on-axis potential of a charged disk from Chapter 25 to...

Question

Use the on-axis potential of a charged disk from Chapter 25 to find the on-axis electric field of a charged disk.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with the relationship between the electric potential and the electric field. So Brown says using the expression for the electric potential at point P located at a perpendicular distance R from a thin wire of length, L with a uniform linear charge density lambda as shown below determine the electric field E at point P. And we're given that VP is equal to LAMBDA divided by the quantity of four pi epsilon. Not that fraction is multiplied by the natural log of the quantity of L divided by two plus the square root of the quantity of R squared plus the quantity of L divided by two in quantity squared. And that's divided by the quantity of minus L divided by two plus the square root of the quantity of R squared plus quantity of L divided by two in quantity squared. Would the question for given a diagram of what was described in the problem? We're also given four possible choices as our answers and we're gonna come back to these answer choices after we've completed the calculation. Now the problem, we were given an equation for the electric potential at this point P. And so we need to use it to calculate our electric field at that point. So recall your relationship between the electric potential and the electric field. So we'll have our electric field E is going to be equal to minus. And here we're gonna start taking partial derivatives will have the partial derivative of V with respect to R. And since I'm taking the partial derivative with respect to a radial direction, that's gonna be multiplied by the R hat unit vector. Now I'd also have two other terms that I would need to look at. Um we'd have the partial of V with respect to theta and the partial of V with respect to Phi but neither of those variables show up in our equation. So when I take those partial derivatives, they're gonna be zero. So that leaves us with only a radio component for our electric field in this case. So at this point, we need to just plug in our formula four V. So we'll have E is equal to minus the partial with respect to R of the quantity of lambda divided by or pi absolutely not multiplied by the natural log of the quantity of L divided by two plus the square root of the quantity of R squared plus quantity of L divided by two quantity squared divided by minus L divided by two plus square root of R squared, LL divided by two quantity squared. That whole expression is still multiplied by our hat. So at this point, I need to just take the derivative will have E is equal to. And here I'm gonna pull the consonant through the derivative. So I have minus lambda divided by four pi epsilon knot. And this is going to be multiplying the quantity. And here I'm taking the derivative of the natural log. So that means I get one divided by the argument. So it's gonna be the minus L divided by two plus the square root of R squared plus L divided by two quantity squared. And that is divided by L divided by two plus the square root of R squared plus the quantity of L divided by two squared. I now need to take the derivative of the argument of the natural log. And here I have a fraction. So I'm going to be using the quotient rule. So for the quotient rule in the denominator, I'm just going to have the denominator squared. So I have the minus L divided by two plus the square root of R squared plus L divided by two quantity squared. And that whole expression is squared. And so now I need to look at the numerator. And for the first term in the numerator, I'm gonna have the derivative of the numerator. So I'll have the derivative of L divided by two, which is just zero. So that term just goes away and then I'll just have the derivative of the square root of R squared plus L divided by two quantity squared. So when I take that derivative, I get R divided by the square root of R squared plus L divided by two quantity squared. And that gets multiplied by the denominator which is minus L divided by two plus the square root of R squared plus L divided by two quantity squared. Then I'll have minus. Now, I'll take the derivative of the denominator. And that happens to be the same as the derivative of the numerator since they only differ by a constant. And so I'll have R divided by the square root of R squared plus L divided by two Y squared. And then that gets multiplied by the new murder which is L divided by two plus the square root of R squared plus L divided by two quantity squared. And that whole expression is still multiplied by our hat. So at this point, I just need to simplify some things. Um I noticed that one of the factors in the numerator cancels out with the, one of the factors in the dominator of the question rule. And so now at this point, I just need to simplify some things. So I'll have E is equal to, I still have my constant, which is the minus lambda divided by or pi epsilon knot. And this is multiplying the quantity. So if I look at my numerator, I'm going to have R divided by the square root of R squared plus L divided by two quantity squared. And that's gonna be multiplying a negative L divided by two and then minus L divided by two. So when I combine those two terms, I just get um minus LR divided by the square root of R squared plus L divided by two quantity squared. And then if I look at the other term, the ones that are the um multiplying the square root of R squared plus L divided by two quantity squared. One of those terms is going to be positive. The other term is gonna be ne negative. So they cancel out. So this is the only term that I have on my numerator. And then this gets divided by my term on the lam leader. And here I just need to multiply um this quantity out. So the first thing I'll have is minus L divided by two quantity squared plus, then I'll have R squared plus L divided by two quantity squared. And this whole thing is still multiplied by our hat. So again, I can still simplify this. So I have E is equal to the minus signs will cancel out. So I'll have a positive lambda divided by four pie. Absolutely not. This is going to get multiplied by. So here um inside the brackets in the denominator, the L divided by two quantity squares will cancel out since one is positive and one's negative. And so that means that also one factor of R will cancel out. And so what I'm left with is L divided by, are multiplied by the square root of R squared plus the quantity of L divided by two quantity squared. And this is still multiplied by our hat. So just sort of rewriting everything together, we're gonna have E is equal to and the numerator will have L multiplied by lambda. This is divided by the quantity of four pi absolutely not multiplied by R multiplying the square root of R squared plus quantity of L divided by two in quantity squared. And that fraction is multiplied by our hat. So this is gonna be our final answer. And when we look at our answer choices, this actually corresponds to answer D. Now if you look at answer C looks very similar, except it has a factor of two in the denominator instead of the factor of four. For choice B, the four pi epsilon knot is actually in the numerator instead of the denominator. And for choice A, we have a factor of two pi epsilon knot in the numerator instead of the four pi epsilon knot in the denominator. So that's why those three answer choices are incorrect. So just a quick little recap of what we've done here. We basically just used our relationship between the electric fields and the electric potentials to calculate our electric field. Given the electric uh potential formula here that consisted of taking partial derivatives in this case with respect to R. So I hope that this has been useful and I'll see you in the next video.

Use the on-axis potential of a charged disk from Chapter 25 to find the on-axis electric field of a charged disk.

Key Concepts

Electric Potential

Electric Field

Gauss's Law

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