Question

I need help with the last part of this problem.

Answer 1

\(\sf F = \dfrac{1}{4 \pi \epsilon}\dfrac{Qq}{x^2} \)

Answer 2

we usually treat x as a radius

Answer 3

here is the question.
A charge is fixed in place. A charge -q is released form rest a distance d from Q. Determine the velocity of -q as a function of its distance x from Q. You will need to write out Newton's 2nd law, apply the chain rule and then integrate both sides of the equation appropriately.
\[\sf F = \dfrac{1}{4 \pi \epsilon}\dfrac{Qq}{d^2}\]

Answer 4

oh okay

Answer 5

2nd law is \(F= ma \) right?

Answer 6

yupp

Answer 7

so try to equate F = ma and the F as charges

Answer 8

is the integral of acceleration a velocity ?

Answer 9

\[\dfrac{1}{4 \pi \epsilon}\dfrac{Qq}{d^2}=m*a\]
\[a=\frac{dv}{dt}\]

Answer 10

Its this right?
\[\frac{F}{m}=a\rightarrow a=\frac{dv}{dt}\]
To find v I need to do integrals, but this is using time. in my function for acceleration I don't have time so I need to do the chain rule? also the question refers to both d and x as distances. I don't know which one to use.

Answer 11

ya we need the velocity

Answer 12

sorry got tied up in the chat there's a party going on

Answer 13

\(F = m \dfrac{dv}{dt} \) does this look better?

Answer 14

i guess

Answer 15

do you have anything better?

Answer 16

not at all.
I'm supposed to used the chain rule:
\[a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v\]

Answer 17

ya that is after the constant factor rule in differentiation
we need to turn m into a constant

Answer 18

\[F = m \dfrac{dv}{dt}\]
\[F=mv\frac{dv}{dx}\]
\[\int Fdx=\int mvdv\]
\[\int Fdx=\frac {mv^{2}}{2}\]

Answer 19

\[\frac{Qq}{4\pi \epsilon_{0}}\int \frac{1}{d^{2}}dx=\frac{mv^{2}}{2}\]
there should be something different in place of the d right? I just don't know what.

Answer 20

i was thinking d-x but that doesn't work

Answer 21

hmm I think this could be fixed a bit