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zubka84 [21]
3 years ago
11

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

nce between the plates is doubled, with the charged particle remaining in the center. Do the force on this particle increase, decrease or stay the same?
Physics
1 answer:
slamgirl [31]3 years ago
4 0

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

C is the capacitance

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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Extend your thinking: household appliances are usually connected in a parallel circuit. why do you think it might be a problem i
alexandr1967 [171]
The equivalent resistance of several devices connected in parallel is given by
\frac{1}{R_{eq}} =  \frac{1}{R_1}+ \frac{1}{R_2}+...+ \frac{1}{R_n}
where R_i are the resistances of the various devices. We can see that every time we add a new device in parallel, the term \frac{1}{R_{eq}} increases, therefore the equivalent resistance of the circuit R_{eq} decreases.

But Ohm's law:
I= \frac{V}{R_{eq}}
tells us that if the equivalent resistance decreases, the total current in the circuit increases. The power dissipated through the circuit (and so, the heat produced) depends on the square of the current:
P=I^2 R
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3 years ago
THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THANKS!!! THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THAN
timurjin [86]

Answer:

f_{o} = 391.67 Hz

Explanation:

The sound of lowest frequency which is produced by a vibrating sting is called its fundamental frequency (f_{o}).

The For a vibrating string, the fundamental frequency (f_{o}) can be determined by:

f_{o} = \frac{v}{2L}

Where v is the speed of waves of the string, and L is the length of the string.

L = 42.0 cm = 0.42 m

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f_{o} = \frac{329}{2*0.42}

   = \frac{329}{0.84}

f_{o} = 391.6667 Hz

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Mrrafil [7]
My answer d absorbed
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