A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
A very small sphere with positive charge 5.00uC is released from rest at a point 1.20cm from a very long line of uniform linear charge density 3.00 uC/m.
What is the kinetic energy of the sphere when it is 4.50cm from the line of charge if the only force on it is the force exerted by the line of charge?
 2 years ago
A very small sphere with positive charge 5.00uC is released from rest at a point 1.20cm from a very long line of uniform linear charge density 3.00 uC/m. What is the kinetic energy of the sphere when it is 4.50cm from the line of charge if the only force on it is the force exerted by the line of charge?

This Question is Closed

yakeyglee
 2 years ago
Best ResponseYou've already chosen the best response.0Let us situate this on the \(x\) axis, and let our uniform line of charge be positioned on the interval \((L, 0]\) for some large number \(L\). The voltage \(V\) as a function of \(x\) on the interval \((0,\infty)\) is given by integrating the contributions from each bit of charge. Let the charge density be \(\lambda\). Thus, for an infinitesimal length element \(dx'\), we have \(\lambda = \frac{dq}{dx'}\).\[V(x) = \frac{1}{4\pi \epsilon_0} \int\limits_{\text{line}} \frac{dq}{r}=\frac{\lambda}{4\pi \epsilon_0} \int\limits_{L}^0 \frac{dx'}{xx'}=\frac{\lambda}{4\pi \epsilon_0} \left( \lnx+L \lnx\right)\]Try using this now to find an approximate voltage drop over your interval. I don't really think you can do it exactly without knowing the value of \(L\).
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.