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dimulka [17.4K]
3 years ago
12

Five air-filled parallel-plate capacitors have the plate areas and plate separations listed below, where A and d are constants.

The capacitors are each connected to the same potential difference. Which capacitor stores the greatest amount of energy?
a.
Area: 2A
Separation : d/2
b.
Area: 2A
Separation : 2d
c.
Area: A
Separation : d
d.
Area: A/2
Separation : d/2
e.
Area: A/2
Separation : 2d
Physics
1 answer:
kirill115 [55]3 years ago
3 0

Answer:

The answer is "Option A"

Explanation:

Energy stored in capcitor=\frac{1}{2}\ cv^2

For point A:

C_A=\frac{\varepsilon_0  2A}{\frac{d}{2}}=\frac{4\ \varepsilon_0 A}{d}\\\\

For point B:

C_B=\frac{\varepsilon_0  2A}{2d}=\frac{\varepsilon_0 A}{d}\\\\

For point C:

C_c=\frac{\varepsilon_0  A}{d}\\\\

For point D:

C_D=\frac{\varepsilon_0  A}{2 \frac{d}{2}}=\frac{\varepsilon_0 A}{d}\\\\

For point E:  C_E=\frac{\varepsilon_0  A}{2 \times 2d}=\frac{\varepsilon_0 A}{4d}\\\\

therefore C_A has the maximum capacitance and max energy same energy is dir proportional to C for the same J

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A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,
alexgriva [62]
The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
S= \frac{1}{2} a t^2
where a is the acceleration. Using the data of the problem, we can find a:
a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J

and so the power output is
P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W

b) power output at t=4.0s 
The velocity at t=4.0 s is
v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(8.0 m/s)^2=1792 J

and so the power output is
P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W

c) Power output at t=6.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J

and so the power output is
P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W
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A small cork with an excess charge of +6.0 μC (1 μC = 10 -6 C) is placed 0.12 m from another cork, which carries a charge of -4.
Alex

Answer:

a.16.125 N b. The force is an attractive force. c. 2.68 × 10¹³ electrons d. 3.75 × 10¹³ electrons

Explanation:

a. What is the magnitude of the electric force between the corks?

The electrostatic force of attraction between the two corks is given by

F = kq₁q₂/r² where k = 9 × 10⁹ Nm²/C², q₁ = +6.0 μC = +6.0 × 10⁻⁶ C, q₂ = -4.3 μC = -4.3 × 10⁻⁶ C and r = distance between the corks = 0.12 m

Substituting the values of the variables into the equation, we have

F = kq₁q₂/r²

F = 9 × 10⁹ Nm²/C² × +6.0 × 10⁻⁶ C × -4.3 × 10⁻⁶ C/(0.12 m)²

= -232.2 × 10⁻³ Nm²/(0.0144 m)²

= -16125 × 10⁻³ N

= -16.125 N

So, the magnitude of the force is 16.125 N

b. Is this force attractive or repulsive?

Since the direction of the force is negative, it is directed towards the positively charged cork, so the force is an attractive force.

c. How many excess electrons are on the negative cork?

Since Q = ne where Q = charge on negative cork = -4.3 μC = -4.3 × 10⁻⁶ C and n = number of excess electrons and e = electron charge = -1.602 × 10⁻¹⁹ C

So n = Q/e = -4.3 × 10⁻⁶ C/-1.602 × 10⁻¹⁹ C = 2.68 × 10¹³ electrons

d. How many electrons has the positive cork lost?

We need to first find the number of excess positive charge n'

Q' = n'q where Q = charge on positive cork = + 6.0 μC = + 6.0 × 10⁻⁶ C and n = number of excess protons and q = proton charge = +1.602 × 10⁻¹⁹ C

So n' = Q'/q = +6.0 × 10⁻⁶ C/+1.602 × 10⁻¹⁹ C = 3.75 × 10¹³ protons

To maintain a positive charge, the number of excess protons equals the number of electrons lost = 3.75 × 10¹³ electrons

4 0
3 years ago
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