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2 years ago

Please answer I need help

Physics

1 answer:

Drupady [299]2 years ago

4 0

Answer:

I have attached a picture of me explaining the answers in written format. Hope this helps, thank you !!

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A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

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3 years ago

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skelet666 [1.2K]

Answer:

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a boy throws a ball straight up into the air. it reaches its highest point after 4 seconds.how fast was the ball going when it l

baherus [9]

Answer:

Explanation:

The most important thing to remember about parabolic motion in physics is that when an object reaches its max height, the velocity right there at the highest point is 0. Use this one-dimensional motion equation to solve this problem:

v = v₀ + at and filling in:

0 = v₀ + (-9.8)(4.0) **I put in 4.0 for time so we have more than just 1 sig fig here**

0 = v₀ - 39 and

-v₀ = -39 so

v₀ = 39 m/s

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gavmur [86]

The answer is A, D (Apex)

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Makovka662 [10]

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Explanation:

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