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3 years ago

How is the potential difference same in capacitors arranged in parallel combination?

Physics

1 answer:

zimovet [89]3 years ago

8 0

Answer:

<h2>Potential difference across capacitors in parallel </h2>

Two or more capacitors are said to be connected in parallel if each one of them is connected across the same two points. In a parallel combination of capacitors potential difference across each capacitor is same but each capacitor will store different charge.

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In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere

jolli1 [7]

1) 333.6 C

In order to have breakdown, the electric field at the surface of the cloud must be equal to the breakdown electric field:

$E=3.00\cdot 10^6 N/C$

The electric field strength at the surface of a charged sphere is given by

$E=\frac{1}{4\pi \epsilon_0} \frac{Q}{R^2}$

where

$\epsilon_0 = 8.85\cdot 10^{-12} C^2/(N^2 m^2)$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

Here we have a cloud of radius

$R = 1.00 km = 1000 m$

So we can re-arrange the previous equation in order to find the charge on the cloud:

$Q=4\pi \epsilon_0 ER^2=4\pi (8.85\cdot 10^{-12})(3.00\cdot 10^6)(1000)^2=333.6 C$

2) $2.08\cdot 10^{21}$ excess electrons

The total charge of the cloud must be (in magnitude)

Q = 333.3 C

We know that one electron carries a charge of

$e = 1.6 \cdot 10^{-19}C$

The total charge is just given by the charge of each electron multiplied by the number of excess electrons in the cloud:

$Q=Ne$

where

N is the number of excess electrons

Solving for N, we find:

$N=\frac{Q}{e}=\frac{333.3 C}{1.6\cdot 10^{-19} C}=2.08\cdot 10^{21}$

6 0

3 years ago

12.0-kg block is pushed across a rough horizontal surface by a force that is angled 30.0◦below thehorizontal. The magnitude of t

drek231 [11]

Answer: Fr = 26.53 N

Explanation: The constant force exerted on the block by the surface is the frictional force.

This frictional force is as a result of interaction between the body and the surface.

According to newton's second law of motion,

F - Fr = ma

F=applied force

Fr = magnitude of frictional force

m = mass of object = 12kg

a = acceleration of object = 3.2m/s²

The applied force (F= 75 N) is inclined at an angle of 30° to the horizontal thus making it have 2 components of forces given below

Fx = 75 * cos 30 = 64.95 N (horizontal component)

Fy = 75 * sin 30 = 37.5 N ( vertical motion)

The body moves across the surface, hence the horizontal component of force is responsible for motion.

F = 64.95 N

By substituting the parameters, we have that

64.96 - Fr = 12 * 3.2

64.96 - Fr = 38.4

Fr = 64.96 - 38.4

Fr = 26.53 N

7 0

3 years ago

Read 2 more answers

A 3.0 g bullet traveling at a speed of 400 m/s enters a tree and exits the other side with a speed of 200 m/s. Where did the bul

cricket20 [7]

Answer:

1800J

Explanation:

Step one:

given data

mass of bullet m= 3g= 0.03kg

initial velocity u = 400m/s

final velocity v= 200 m/s

Step two:

1.The bullet's lost kinetic energy went inside the tree.

2. The energy transferred is computed as

= initial KE- KE final

Initial KE= 1/2mu^2

Initial KE= 1/2*0.03*400^2

Initial KE= 1/2*0.03*160000

Initial KE= 1/2*4800

Initial KE= 2400J

KE final= 1/2mv^2

KE final= 1/2*0.03*200^2

KE final= 1/2*0.03*40000

KE final= 1/2*1200

KE final= 600J

KE transferred = 2400-600

KE transferred= 1800J

6 0

3 years ago

It takes 56.5 kilojoules of energy to raise the temperature of 150 milliliters of water from 5°C to 95°C. If you

natka813 [3]

If 56.5kJ are needed to raise the temp by 90°C and if the heater is 60% efficient that means that:
60% X y = 56.5kJ
where y is the electrical energy in kJ that the heater will use.
y = 94.2kJ

6 0

3 years ago

A string attached to an airborne kite is maintained at an angle of 41 degrees with the horizontal. If a total of 152 m of string

satela [25.4K]

Answer:

The horizontal displacement is $Adj = 114.71 \ m$