Physics, 13.12.2019 18:31 badpotterchris

Aspherical capacitor contains a charge of 3.50 nc when connected to a potential difference of 230.0 v. its plates are separated by vacuum and the inner radius of the outer shell is 4.30 cm.

a) calculate the capacitance.
b) calculate the radius of the inner sphere.
c) calculate the electric field just outside the surface of theinner sphere.

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Answer from: ally7366

(A). The capacitance is $15.2\times10^{-12}\ F$

(B). The radius of the inner sphere is 0.0327 m .

(C). The electric field just outside the surface of the inner sphere is $2.9458\times10^{4}\ V/m$

Explanation:

Given that,

Charge = 3.50 nC

Potential difference = 230.0 V

Radius = 4.30 cm

(a). We need to calculate the capacitance

Using formula of capacitance

$C=\dfrac{Q}{V}$

$C=\dfrac{3.50\times10^{-9}}{230.0}$

$C=15.2\times10^{-12}\ F$

(b). We need to calculate the radius of the inner sphere

Using formula of capacitance

$C=\dfrac{4\pi\epsilon_{0}r_{a}r_{b}}{r_{b}-r_{a}}$

Put the value into the formula

$15.2\times10^{-12}=4\pi\times8.85\times10^{-12}(\dfrac{r_{a}\times4.30\times10^{-2}}{4.30\times10^{-2}-r_{a}})$

$\dfrac{r_{a}4.30\times10^{-2}}{4.30\times10^{-2}-r_{a}}=\dfrac{15.2\times10^{-12}}{4\pi\times8.85\times10^{-12}}$

$\dfrac{r_{a}4.30\times10^{-2}}{4.30\times10^{-2}-r_{a}}=0.13667$

$r_{a}\times4.30\times10^{-2}=0.13667\times(4.30\times10^{-2}-r_{a})$

$r_{a}(4.30\times10^{-2}+0.13667)=0.13667\times4.30\times10^{-2}$

$r_{a}=\dfrac{0.13667\times4.30\times10^{-2}}{(4.30\times10^{-2}+0.13667)}$

$r_{a}=0.0327\ m$

(c). We need to calculate the electric field just outside the surface of the inner sphere

Using formula of electric filed

$E=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{Q}{r_{a}^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times3.50\times10^{-9}}{0.0327}$

$E=2.9458\times10^{4}\ V/m$

Hence, (A). The capacitance is $15.2\times10^{-12}\ F$

(B). The radius of the inner sphere is 0.0327 m .

(C). The electric field just outside the surface of the inner sphere is $2.9458\times10^{4}\ V/m$

Answer from: Quest

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .

Look at the pendulum diagram between which points does the pendulum’s potential energy increase most

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.

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