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2 years ago

A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.

1 answer:

7 0

The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as

t=0.476v
t=1.967v
V2=4.323v

<h3>What is the potential across the capacitor?</h3>

Question Parameters:

A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.

t = 1.0 seconds
5.0 seconds
20.0 seconds.

Generally, the equation for the Voltage is mathematically given as

v(t)=Vmax=(i-e^{-t/t})

Therefore

For t=1

V=5(i-e^{-1/10})

t=0.476v

For t=5s

V2=5(i-e^{-5/10})

t=1.967

For t=20s

V2=5(i-e^{-20/10})

V2=4.323v

Therefore, the values of voltages at the various times are

t=0.476v
t=1.967v
V2=4.323v

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Answer:

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3 years ago

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3 years ago

An insulating hollow sphere has inner radius a and outer radius

aleksklad [387]

The solution can be given using the Gauss' Theorem. First let's make the following assumptions:
*This problem has a perfect spherical symmetry so the field for a charge density of $\rho(r)$ that depends only on the radial component(we are of course working in spherical coordinates) will yield a constant electric field pointing radially outwards,that is, the field will be uniform for a<r<b independently of the angles $\theta$ and $\phi$

*As a result of the uniformity of the electric field, E will be constant and it can be taken out of the integral involving Gauss' law.
a)
Now let's get our hands to it, recall that Gauss' Law states:
$\oint\mathbf{E}.d\mathbf{a}=\frac{Q_{enclosed}}{\epsilon_0} \\ \impliesE\oint da=E.4\pi r^2=\frac{Q_{enclosed}}{\epsilon_0}$
Here we exploited the justified assumption that The field is uniform, because the field is pointing in the outward radial direction the scalar product between the field and surface element will yield $\mathbf{E}.d\mathbf{a}=Edacos(0)=Eda$
Now we need to determine the enclosed charge: $Q_{enclosed}=\int_V\rho(r)dV$
In spherical coordinates we thus have:
$Q_{enclosed}=\frac{1}{\epsilon_0}\int_V\rho(r)dV=\int^b_a\int^{2\pi}_0\int^{\pi}_0\frac{\alpha}{r}r^2sin\theta drd\theta d\phi$ phi[/tex] over all of the gaussian surface.
Where the integral:
$\int^{2\pi}_0\int^{\pi}_0sin\theta d\theta d\phi=4\pi$
Returning to our integral we have:

$\frac{1}{\epsilon_0}\int^b_a\int^{2\pi}_0\int^{\pi}_0\frac{\alpha}{r}r^2sin\theta drd\theta d\phi=\frac{1}{\epsilon_0}\int^a_b\frac{\alpha}{r}r^2dr\int^{2\pi}_0\int^{\pi}_0sin\theta d\theta d\phi=\frac{1}{\epsilon_0}4\pi \int^b_a\alpha rdr
\\
\\
=\frac{1}{\epsilon_0}4\pi\alpha\left[\frac{1}{2}r^2\right]^b_a=\frac{1}{\epsilon_0}2\pi\alpha(b^2-a^2)$
Now it's just a matter of solving for E:
$E=\frac{\alpha}{2\epsilon_0}\frac{b^2-a^2}{r^2}$
b)
If a point charge is placed at the center of our system the the resulting field will be the sum of both fields, this field needs to be constant, let's pick for now that the field is zero in the region a<r<b:
$E_p+E=0$
Where $E_p$ is the field due to a point charge.
Again using Gauss's theorem the field of a point charge 1 is:
$E_p\oint da=E.4\pi r^2=\frac{Q_{enclosed}}{\epsilon_0}=\frac{q}{\epsilon_0}
\\
\\
\implies E_p=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$
We then get the following expression:
$E_p+E=0
\\
\implies E=\frac{\alpha}{2\epsilon_0}\frac{b^2-a^2}{r^2} +\frac{1}{4\pi\epsilon_0}\frac{q}{r^2} =0$
Solving for q we get:
$q=-2\pi\alpha(b^2-a^2)$

If instead we want a non zero field $E=c$ then we only need to solve
$E_p+E=c$ which yields:
$E_p+E=c \\ \implies \frac{\alpha}{2\epsilon_0}\frac{b^2-a^2}{r^2} +\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}=c
\\
\\
\implies q=-2\pi[\alpha(b^2-a^2)-c]$

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3 years ago

Resultant of a pair of 100 km/h velocity vectors that are at right angels to each other

nignag [31]

The resultant of the two vectors is 141 km/h.

Explanation:

When two vectors are at right angles to each other, their resultant can be calculated by using the Pythagorean's theorem:

$R=\sqrt{A^2+B^2}$