Physics, 14.02.2020 03:29 jj1077348

An empty plate capacitor is connected between the terminals ofa 9.0 V battery and charged up. The capacitor is then disconnectedfrom the battery, and the spacing between the capacitor plates isdoubled. As a result of the change, what is the new voltage betweenthe plates of the capacitor?
I have:
E = v/d
E =9 V
then the distance is doubled 9(2) = 18 V
I feel like maybe I over simplified the problem. Does thislook right?

Answers: 1

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The new voltage between the parallel plates of the capacitor is 18V, because for a constant electric field, doubling the space between the parallel capacitor plates, will also double the potential difference (voltage) between the plates.

Explanation:

ΔV = E*Δd

Where;

ΔV is the change in potential difference

Δd is the change in the distance between the parallel plates

E is the electric field potential.

Assuming a constant electric field; $E = \frac{v}{d} , then; \frac{v_1}{d_1} =\frac{v_2}{d_2}$

when the spacing between the capacitor plates is doubled, d₂ = 2d₁

v₂ = (v₁*d₂)/(d₁)

v₂ = (v₁*2d₁)/(d₁)

v₂ = 2v₁

v₂ = 2(9) = 18 V

Therefore, for a constant electric field, doubling the space between the parallel capacitor plates, will also double the potential difference (voltage).

Answer from: Quest

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.

Answer from: Quest

Amicroscope, because these cells are so small that you would not be able to see them, or what they attack without it.

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