This really helps, thanks for making it available. However, I am wondering why the denominator is not just s-squared since the problem states that we want to know the e-field a distance s from the end of the rod.
Because not all the charge is located at the end of the rod! You have to account for the varying distance between the charge contributions and the observation point by using a physical integral, and most of those charge contributions lie at a distance greater than s from the observation point. By the way, at the end of the video we also look at the large s limit of the result, and we see that the field reduces to a 1/s^2 point-charge type field when we are very far from the rod, so that's where you find the 1/s^2 behavior you're familiar with for point charges. z
This is a really good question, and I'm likely to use it some day as a bonus question on an exam for my first year engineering physics students! My first idea would be to chop the cylinder into thin disks, then use the previous result for the electric field on the axis of a thin disk, which you can find here: ua-cam.com/video/8yw1_jhjuHQ/v-deo.html (this solution is in turn built on the result for the electric field of a thin ring on axis, so your question is kind of the natural extension of building new results from previous results). -- Zak
🎉thank you for explaining for us the way we can apply to get the electric field
I really love this information.. please make integration techniques easier and easier
You do a great job at explaining, thank you. Hope i can remember this for my test .🤣
thanks! z
This really helps, thanks for making it available. However, I am wondering why the denominator is not just s-squared since the problem states that we want to know the e-field a distance s from the end of the rod.
Because not all the charge is located at the end of the rod! You have to account for the varying distance between the charge contributions and the observation point by using a physical integral, and most of those charge contributions lie at a distance greater than s from the observation point. By the way, at the end of the video we also look at the large s limit of the result, and we see that the field reduces to a 1/s^2 point-charge type field when we are very far from the rod, so that's where you find the 1/s^2 behavior you're familiar with for point charges. z
@@ZaksLab Thank you very much for taking the time to explain further. I appreciate this.
what if you have a cylinder instead of a rod ?
This is a really good question, and I'm likely to use it some day as a bonus question on an exam for my first year engineering physics students! My first idea would be to chop the cylinder into thin disks, then use the previous result for the electric field on the axis of a thin disk, which you can find here: ua-cam.com/video/8yw1_jhjuHQ/v-deo.html (this solution is in turn built on the result for the electric field of a thin ring on axis, so your question is kind of the natural extension of building new results from previous results). -- Zak
Ahhhh, so simple