All categories

2 years ago

What is the electric potential, i.e. the voltage, 0.30 m from a point charge of 6.4 x 10-C?

Physics

1 answer:

Gwar [14]2 years ago

3 0

Answer:

V = 192 kV

Explanation:

Given that,

Charge, $q=6.4\times 10^{-6}\ C$

Distance, r = 0.3 m

We need to find the electric potential at a distance of 0.3 m from a point charge. The formula for electric potential is given by :

$V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 6.4\times 10^{-6}}{0.3}\\\\V=192000\ V\\\\V=192\ kV$

So, the required electric potential is 192 kV.

You might be interested in

Analogue signals transmit information for such things as _____________.

ivann1987 [24]

Transmission of information in ANY form can be done digitally
or analoguely.

Beginning about 30 years ago, everything slowly started changing
to digital. Today, all commercial satellite communication, all optical
fiber communication, all internet communication, all computer
communication, all commercial cable communication, all commercial
television, and much of the telephone system, are all digital.

On your computer ... .pdf, .jpg, .mp3 etc. are all digital methods of
moving and storing information.

AM and FM radio are an interesting subject. They're all still analog.
They could easily be changed to all digital, and it would be a big
improvement, both for the broadcasters and for the listeners.
BUT ... every AM and FM radio that anybody has now would be
obsolete. Every single radio would either need to be replaced,
OR you'd need to add a digital decoder to every radio, like we
had to do with our TV sets a few years ago when television
suddenly became all digital. With AM and FM radios, the decoders
would be bigger, and would cost more, than most of the radios.

And that's why commercial radio broadcasting is still analog.

7 0

2 years ago

Read 2 more answers

What is the sun's thickest layer?<br>besides radiation zone?

makvit [3.9K]

Corona gas star is literally z thickest layer. hope it helped.

3 0

2 years ago

Read 2 more answers

An object is located 51 millimeter from a diverging lens. te object has a hight of 13 millimeters and the image height is 3.5 mi

Sati [7]

<span>An object is located 51 millimeters from a diverging lens the object has a height og 13 millimeters and the image height is 3.5 millimeters how far in front of the lens is image located?</span>

3 0

2 years ago

"what shape are the pebble-sized particles that make up the rock?"

Fiesta28 [93]

The shape of which the pebble sized particles that makes up the rock is more of a rounded shape as they are bits and in circular formation, depicting it to be more rounder in which are particles that is made up in a rock. They can be in pebble sized or much more smaller than that when in bits.

5 0

2 years ago

How long does it take electrons to get from

OlgaM077 [116]

We need to find the time it takes an electron to move in the given circuit.

The time taken for electrons to reach the starting motor from the battery is 60.65 minutes.

I = Current = 134 A

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

A = Area = $38.9\ \text{mm}^2$

L = Length = 92.2 cm

$\rho$ = Density of copper = $8960\ \text{kg/m}^3$

M = Molar mass of copper = 63.5 g/mol

$n_v$ = Number of valence electrons of copper = 1

e = Charge of electron = $1.6\times 10^{-19}\ \text{C}$

Number of charge carriers per unit volume is given by

$n=\dfrac{\rho N_An_v}{M}\\\Rightarrow n=\dfrac{8960\times 6.022\times 10^{23}\times 1}{63.5\times 10^{-3}}\\\Rightarrow n=8.497\times 10^{28}\ \text{m}^{-3}$

Time taken is given by

$t=\dfrac{LAne}{I}\\\Rightarrow t=\dfrac{92.2\times 10^{-2}\times 38.9\times 10^{-6}\times 8.497\times 10^{28}\times 1.6\times 10^{-19}}{134}=3638.83\ \text{s}\\\Rightarrow t=\dfrac{3638.83}{60}=60.65\ \text{minutes}$

The time taken for electrons to reach the starting motor from the battery is 60.65 minutes.

Learn more:

brainly.com/question/1426683

brainly.com/question/170663

8 0

1 year ago