Ask question

Ask question

All categories

babymother [125]

3 years ago

7

Two plates of area 7.00 × 10-3 m2 are separated by a distance of 2.60 × 10-4 m. If a charge of 5.40 × 10-8 C is moved from one p

late to the other, calculate the potential difference (voltage) between the two plates. Assume that the separation distance is small in comparison to the diameter of the plates.

1 answer:

lesantik [10]3 years ago

6 0

Answer:

226.53 Volt

Explanation:

A = Area of plates = 7.00×10⁻³ m²

ε₀ = Permittivity of space = 8.854×10⁻¹² F/m

d = Distance between two plates = 2.60×10⁻⁴ m

Q = Charge = 5.40×10⁻⁸ C

Capacitance

$C=\frac{\epsilon_0A}{d}\\\Rightarrow C=\frac{8.854\times 10^{-12}\times 7\times 10^{-3}}{2.6\times 10^{-4}}\\\Rightarrow C=23.83\times 10^{-11}$

Potential difference between plates

$V=\frac{Q}{C}\\\Rightarrow V=\frac{5.4\times 10^{-8}}{23.83\times 10^{-11}}\\\Rightarrow V=226.53\ Volt$

∴ The potential difference (voltage) between the two plates is 226.53 Volt

You might be interested in

A ball is dropped from a height of 10m. At the same time, another ball is thrown

soldi70 [24.7K]

5.1 m

Explanation:

Let's set the ground as our reference point. Let's also call the dropped ball to be ball #1 and its height above the ground at any time t is given by

$y_1 = 10 - \frac{1}{2}gt^2$ (1)

where 10 represents its initial height or displacement of 10 m above the ground. At the same time, the displacement of the second ball with respect to the ground $y_2,$ is given by

$y_2 = v_0t - \frac{1}{2}gt^2$ (2)

At the instant the two balls collide, they will have the same displacement, therefore

$y_1 = y_2 \Rightarrow 10 - \frac{1}{2}gt^2 = v_0t - \frac{1}{2}gt^2$

or

$v_0t = 10\:\text{m}$

Solving for t, we get

$t = \dfrac{10\:\text{m}}{v_0} = \dfrac{10\:\text{m}}{10\:\text{m/s}} = 1\:\text{s}$

We can use either Eqn(1) or Eqn(2) to hind the height where they collide. Let's use Eqn(1):

$y_1 = 10\:\text{m} - \frac{1}{2}(9.8\:\text{m/s}^2)(1\:\text{s})^2$

$\:\:\:\:\:\:\:= 5.1\:\text{m}$

8 0

3 years ago

Which phrases describe all the outer planets’ motion? Check all that apply.

uranmaximum [27]

Answer: slow revolution and fast rotation

Solar system has 8 planets. 4 inner rocky planets - Mercury, Venus, Earth and Mars and 4 outer gaseous planets - <u>Jupiter, Saturn, Uranus and Neptune.</u> The outer planets have few common features.

They are gaseous. There period of revolution is larger than the inner planets which means that they have slow revolution about the Sun. One day on the outer planets is smaller than the inner planets which means they have fast rotation.

<u>For example,</u> Jupiter has revolves around sun in 11.86 Earth years and rotates about axis in 9.8 Earth hours. Uranus revolves around sun in 84 Earth years and rotates on its axis 17.9 Earth hours.

8 0

3 years ago

Read 2 more answers

Acceleration problem <br> Show work plz

Dennis_Churaev [7]

Answer:

The answer to your question is: vo = 25 m/s

Explanation:

data

a = -7.5 m/s²

d = 42 m

vf = 0 m/s

vo = ?

Formula

vf² = vo² - 2ad

Substitution

0² = vo² - 2(7.5)(42)

We clear vo from the equation

vo² = 2(7.5)(42)

vo² = 630 simplifying

vo = 25 m/s result

3 0

3 years ago

NASA scientists suggest using rotating cylindrical spacecraft to replicate gravity while in a weightless environment. Consider s

dusya [7]

Answer:

Explanation:

Given

diameter of spacecraft $d=148\ m$

radius $r=74\ m$

Force of gravity $F_g$ =mg

where m =mass of object

g=acceleration due to gravity on earth

Suppose v is the speed at which spacecraft is rotating so a net centripetal acceleration is acting on spacecraft which is given by

$F_c=\frac{mv^2}{r}$

$F_c=F_g$

$\frac{mv^2}{r}=mg$

$\frac{v^2}{r}=g$

$v=\sqrt{gr}$

$v=\sqrt{1450.4}$

$v=38.08\ m/s$

8 0

3 years ago

A block of 200 g is attached to a light spring with a force constant of 5 N / m and freely in a horizontal plane vibrates. The m

Finger [1]

Answer:

m  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 mm  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 m

Explanation:

That is a reason

8 0

3 years ago