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Rama09 [41]
3 years ago
10

Two parallel plate capacitors 1 and 2 are identical except that capacitor 1 has charge +q on one plate and charge −q on the othe

r, and capacitor 2 has charge +2q on one plate and −2q on the other. Compare these two capacitors based on their (a) capacitance, (b) potential difference between the plates, (c) electric field magnitude between the plates,(d) energy stored.
Physics
1 answer:
Grace [21]3 years ago
4 0

Answer:

a) the capacitance is the same for both capacitors.

b) The potential difference between the plates for the capacitor with charge +2q, is double of the one for the capacitor  with charge +q.

c) The electric field magnitude between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q

d) The energy stored between the plates for the capacitor with charge +2q, is 4 times the value for the one with charge +q  

Explanation:

a) The capacitance of a capacitor, by definition, is as follows:

C = \frac{q}{V}

Appying Gauss' Law to one of plates, it can be showed, that the capacitance (for a parallel plates capacitor) can be  expressed as follows:

C = ε*A / d

As it can be seen, it does not depend on the charge. so we conclude that the capacitance must be the same for both capacitors, due to they are identical except for the value of the charge on the plates.

b) By definition, as we said above, the capacitance is equal to the proportion between the charge of one of the plates, and the potential difference between them.

If this proportion must remain the same, and one of the capacitors has the double of  the charge than the other, the potential difference must be the double also.

c) Applying Gauss' law, to the surface of one of  the plates, and assuming a constant surface charge density σ, it can be  showed that the  electric field can be calculated as follows:

E*A = Q/ε₀ as σ=Q/A

⇒ E = σ/ε₀

As σ is directly proportional to the charge (being the area A the same), we conclude that the electric field for the capacitor with charge +2q must be the double than the one for the capacitior with charge +q.

c) The electric potential energy, stored between plates of a capacitor, can be written as follows:

Ue = \frac{1}{2} *\frac{q^{2}}{C}

If the capacitance remains the same, we can conclude that the electric potential energy for the capacitor with charge +2q, as the charge is raised to the 2nd power, must be 4 times the one for the capacitor with charge +q.

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In what fundamental way did the work of Galileo differ from his predecessors who had thought about the sky?
kompoz [17]

Answer: The correct answer is:

Galileo used instruments and experiments to show him what nature was doing, instead of relying on pure logic

Explanation:

Galileo can be considered as one of the precursors of experimentation and the scientific method. A method that doesn't rely on "common sense" and rationalization and logic, but instead is fuelled by a disposition of skepticism and rather makes claims about reality based on experimentation and empirical data shows.

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3 years ago
A 125-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a
lbvjy [14]

Answer:

201.6 N

Explanation:

m = mass of disk shaped merry-go-round = 125 kg

r = radius of the disk = 1.50 m

w₀ = Initial angular speed = 0 rad/s

w = final angular speed = 0.700 rev/s = (0.700) (2π) rad/s = 4.296 rad/s

t = time interval = 2 s

α = Angular acceleration

Using the equation

w = w₀ + α t

4.296 = 0 + 2α

α = 2.15 rad/s²

I = moment of inertia of merry-go-round

Moment of inertia of merry-go-round is given as  

I = (0.5) m r² = (0.5) (125) (1.50)² = 140.625 kgm²

F = constant force applied

Torque equation for the merry-go-round is given as

r F = I α

(1.50) F = (140.625) (2.15)

F = 201.6 N

4 0
3 years ago
A 2 kg ball if at rest. If the ball accelerates to 20 m/s, what is the change in momentum?
katrin [286]

Answer:

change in momentum

20m/s x2=40kg/m

4 0
3 years ago
In a race, Usain Bolt accelerates at
jeka94

Answer:

65.87 s

Explanation:

For the first time,

Applying

v² = u²+2as.............. Equation 1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance

From the question,

Given:  u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m

Substitute these values into equation 1

v² = 0²+2(1.99)(60)

v² = 238.8

v = √238.8

v = 15.45 m/s

Therefore, time taken for the first 60 m is

t = (v-u)/a............ Equation 2

t = (15.45-0)/1.99

t = 7.77 s

For the final 40 meter,

t = (v-u)/a

Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²

Substitute into the equation above

t = (0-15.45)/-0.266

t = 58.1 seconds

Hence total time taken to cover the distance

T = 7.77+58.1

T = 65.87 s

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3 years ago
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