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3 years ago

A 4V source is connected across a 2μF capacitor. Find the charge on this capacitor.

Physics

1 answer:

Feliz [49]3 years ago

6 0

Answer: $8$ μc

have: 2μF = $2.10^{-6}$ F

the charge on this capacitor is $2.10^{-6}.4=8.10^{-6}(c)=8$ (μc)

Explanation:

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3 years ago

The gravitational force of attraction between two students sitting at their desks in physics class is 2.59 × 10−8 N. If one stud

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<h2>The distance between students is 2.46 m</h2>

Explanation:

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3 years ago

Read 2 more answers

A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. SUppose the dista

slamgirl [31]

Answer:

Stay the same

Explanation:

First of all, let's find how the capacitance of the capacitor changes.

Initially, it is given by

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:

$C' = \frac{1}{2}C$

The potential difference across the capacitor is given by

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

where

Q is the charge on the plates

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We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:

$V' = 2 V$

Now we can analyze the electric field between the plates of the capacitor, which is given by

$E=\frac{V}{d}$

we said that:

- The voltage has doubled: V' = 2 V

- The distance between the plates has doubled: d' = 2 d

therefore, the new electric field will be

$E'=\frac{2V}{2d}=\frac{V}{d}=E$

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

F = qE

Then the force on the particle will stay the same.

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3 years ago

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

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Answer:

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Explanation:

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$y(t) = h + v_y t - \frac{1}{2}gt^2$