Physics, 20.09.2020 19:01, kharmaculpepper

What is VB, the electric potential at point B, located at distance d from one end of the rod (on the x axis)?

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answered: jasminecoronetti44

The question is missing some parts. Here is the complete question.

A finite rod of length L has a total charge q, distributed uniformly along its length. The rod lies on the x-axis and is centered at the origin. Thus one endpoint is located at (L/2) and the other is located at (L/2). Define the electrical potential to be zero at an infinite distance away from the rod. Throughout this problem, you may use the constant k in place of the expression 1 / $4\pi\epsilon_{0}$

Part A (image 1): What is VA, the electric potential at point A (see figure below), is located a distance d above the midpoint of the rod on the y-axis? Express your answer in terms of L, d, q and k.

Part B (image 2): What is VB, the electric potential at point B, located at distance d form one end of the rod (on the x-axis)? give your answer in terms of L, d, q and k.

Part A: $V = \frac{kq}{L}[ln(\frac{L/2+\sqrt{(L/2)^{2}+y^{2}} }{-L/2+\sqrt{(-L/2)^{2}+y^{2}} }) ]$

Part B: $V=\frac{kq}{L} ln(\frac{L}{d} )$

Explanation: Electric Potential (V) is the amount of work done per unit charge to move a charge from point A to B.

For a finite rod with charge uniformly distributed along its length

$V=\frac{1}{4\pi\epsilon_{0}} \int\ {\frac{\lambda}{r} } \,dl$

where

λ is charge density and, in this case, is constant: $\lambda=\frac{q}{L}$

dl is differential of the rod

r is the distance the point is from the rod

Part A: $r = \sqrt{(a^{2}+y^{2})}$

$V=k \int\ {\frac{\lambda}{\sqrt{(a^{2}+y^{2})}} } \,da$

$V=k\lambda \int\ {\frac{1}{{\sqrt{(a^{2}+y^{2})}}} } } \,da$

$V=\frac{kq}{L}\int\limits^\frac{L}{2} _\frac{-L}{2} {\frac{1}{\sqrt{(a^{2}+y^{2})} } } \, da$

$V=\frac{kq}{L}ln(\frac{L/2+\sqrt{(L/2)^{2}+y^{2}} }{-L/2+\sqrt{(-L/2)^{2}+y^{2}} } )$

At a point located at y-axis, electric potential is $V=\frac{kq}{L}ln(\frac{L/2+\sqrt{(L/2)^{2}+y^{2}} }{-L/2+\sqrt{(-L/2)^{2}+y^{2}} } )$

Part B: r = x

$V=k \int\limits^L_d {\frac{\lambda}{x} } \, dx$

$V=k \frac{q}{L} \int\limits^L_d {\frac{1}{x} } \, dx$

$V=k \frac{q}{L} ln(L - d)$

$V= \frac{kq}{L} ln(\frac{L}{d} )$

At a point d from one end, electric potential is $V= \frac{kq}{L} ln(\frac{L}{d} )$

What is VB, the electric potential at point B, located at distance d from one end of the rod (on the

answered: Guest

f is prop to m1xm2 double m1 double force

answered: Guest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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