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ahrayia [7]
3 years ago
9

A parallel plate capacitor is charged up by a battery. The battery is then disconnected, but the charge remains on the plates. T

he plates are then pulled apart. Explain whether each of the following quantities increases, decreases, or remains the same as the distance between the plates increases.
(a) The capacitance of the capacitor
(b) The potential difference between the plates
(c) The electric field between the plates
(d) The electric potential energy stored by the capacitor.
Physics
1 answer:
Mazyrski [523]3 years ago
3 0

Given a capacitor that is charge by a battery

So when the battery was disconnect, the charge remains on the capacitor

One plate will have positive charge

And the other will have a negative charge but of same magnitude

Now, the plates are pulled apart, so the distance between them was extended

a. The capacitance of the capacitor?

The capacitance of a capacitor is given as

C = εo•A/d

Where C is the capacitance

εo is permissivity, a constant

A is are of plate

d is the distance apart

From the formula we notice that

Capacitance is inversely proportional to distance, so this implies that as the distance is increase, the capacitance is reduce.

Capacitance Decreases

b. Potential difference?

The potential difference is given as

q = CV

V = q/C

Or V=Ed

Where E is electric field

So the potential difference is inversely proportional to the capacitance, so we know that the capacitance is reducing, then voltage will increase.

Or, using the second relations

V=Ed, voltage is directly proportional to distance, So, if the distance apart is increasing then, the voltage is increasing

Voltage is increasing

c. Electric field between plate.

The electric field is given as

E = kQ/r²

So the electric field is inversely proportional to the distance square, so as the distance is an increase, the electric field is decreased.

Electric field decreases

d. Electric potential energy?

U = ½CV², then, q = CV

Then, U = ½qV

The electric potential energy is directly proportional to the voltage and since the voltage is increasing due to increase in distance, then the electric potential energy increases.

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Substituting, we find:

v=(3.14)(0.50)=1.57 m/s

B) Faster

In this case, the drum is twice as wide, so the new radius of the circular path of the sock is twice the previous one:

r' = 2r = 1.00 m

At the same time, the drum spins at the same frequency as before, therefore the angular frequency as not changed:

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Therefore, the new linear speed would be:

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Two identical 7.10-gg metal spheres (small enough to be treated as particles) are hung from separate 700-mmmm strings attached t
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Answer:

Explanation:

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Let T be tension in the hanging string

T cosθ = mg ( for balancing in vertical direction )

for balancing in horizontal direction

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Dividing the two equations

Tanθ = F / mg

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F = 21.27 x 10⁻³ N

if q be charge on each sphere , force of repulsion between the two

F = k q x q / r² ( r is distance between two sphere , r = 2 x .7 x sin17  = .41 m )

21.27 x 10⁻³  = (9 X 10⁹ x q²) / .41²

q² = .3973 x 10⁻¹²

q = .63 x 10⁻⁶ C

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Novosadov [1.4K]

Answer:

A) correct answer is C,   B)   correct answer is b  and C) The correct answer is b

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In the exercises of geometric optics, the equation of the constructor tells us the location of the image.

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where f is the focal length of the cornea-crystalline system, p and q are the distances to the object and the image.

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