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nikklg [1K]
3 years ago
6

The capacitance of an isolated charged sphere is independent of both the ______ on the sphere and the __________, it is dependen

t only on its_______​
Physics
1 answer:
Anestetic [448]3 years ago
6 0

The capacitance of an isolated charged sphere is independent of both the <u>charge </u>on the sphere and the<u> potential difference</u>, it is dependent only on its <u>radius</u>

The capacitance of an isolated sphere is calculated as follows;

  • let the charge on the sphere = Q
  • let the potential difference = V
  • let the radius of the sphere = R

The potential difference is given as;

V = \frac{kQ}{R}

where;

k is Coulomb's constant = 4\pi \epsilon _0

The capacitance is given as;

c = \frac{Q}{\Delta V} = \frac{Q}{V_1 - V_2} \\\\at \ infinite \ radius \ R , V_2 = 0\\\\c = \frac{Q}{V_1 } \ \ \ \ (V_1 = V = \frac{kQ}{R} )\\\\c = \frac{Q}{kQ/R} \\\\c = \frac{R}{k}\\\\c = \frac{R}{4\pi \epsilon_0}

From the equation above, the capacitance (c) is directly proportional to its radius (R) and independent of both the charge (Q) on the sphere and the potential difference (V).

Thus, the capacitance of an isolated charged sphere is independent of both the <u>charge </u>on the sphere and the<u> potential difference</u>, it is dependent only on its <u>radius</u>

Learn more here: brainly.com/question/12948250

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Answer:

461 km/h

Explanation:

In order to solve this problem we must first sketch a drawing of what the situation looks like so we can better visualize it. (See attached picture).

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The second situation is the police that is accelerating from rest until he reaches Mustapha. (We are going to suppose the acceleration is constant and that he will not stop accelerating until he reaches Mustapha). He has an acceleration of 15m/s^{2}.

We want to find what the final velocity of the police is at the time he reaches Mustapha. From this we can imply that the displacement x will be the same for both particles.

So let's model the first situation.

The displacement of Mustapha can be found by using the following equation:

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when solving the equation for the displacement x we get that it will be:

x=V_{M}t

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x=V_{0}t+\frac{1}{2}at^{2}

Since the initial velocity of the cop is zero, we can get rid of that part of the equation leaving us with:

x=\frac{1}{2}at^{2}

We can now set both equations equal to each other so we get:

\frac{1}{2}at^{2}=V_{M}t

When solving this for t, we get that:

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Now, we know the cop has constant acceleration, so we can model it with the following formula too:

a=\frac{V_{f}-V_{0}}{t}

since the initial velocity of the cop is zero, we can get rid of that here too, so we get the following formula:

a=\frac{V_{f}}{t}

when solving for the final velocity, we get that:

V_{f}=at  (let's call this equation 2)

when substituting equation 1 into equation 2 we get:

V_{f}=a(\frac{2V_{M}}{a})

we can now cancel a leaving us with:

V_{f}=2V_{M}

This tells us that the final velocity of the cop will not depend on his acceleration. (This is only if the acceleration is constant all the time) So we get that the final velocity of the cop is:

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so

V_{f}=461km/hr

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Answer:

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