Electric Potential expression at point P

Thread starter ashworcp
Start date May 18, 2011
Tags

Electric Electric potential Expression Point Potential

AI Thread Summary

To find the electric potential at point P on the y-axis due to a plastic rod with a nonuniform linear charge distribution λ = cx, the relevant equation is V = 1/4πε ∫(λ(x) dx)/r, where r is the distance from the charge element to point P. The charge distribution varies with position along the rod, necessitating integration across its length. The integration limits will depend on the rod's length L and the distance y from point P. The final expression for the electric potential V can be derived by evaluating this integral, considering the geometry of the setup.

May 18, 2011

ashworcp

Homework Statement

The plastic rod of length L shown in the figure has the nonuniform linear charge distribution λ =cx, where c is a positive constant. Find an expression for the electric potential at point P on the y-axis, a distance y from one end of the rod.

V = ?

Homework Equations

V = 1/4πϵ

Physics news on Phys.org

May 19, 2011

rl.bhat

Homework Helper

There is no figure in the problem.

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...

View full post »

Thread 'Explain this asphalt sheen'

This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is The second part of the problem is exactly why you get this affect. And one more spoiler:

View full post »

Thread 'A cylinder connected to a hanged mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

View full post »