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katrin2010 [14]

3 years ago

5

A simple model of a hydrogen atom is a positive point charge +e (representing the proton) at the center of a ring of radius a wi

th negative charge −e distributed uniformly around the ring (representing the electron in orbit around the proton). Find the magnitude of the total electric field due to this charge distribution at a point a distance a from the proton and perpendicular to the plane of the ring.

1 answer:

Norma-Jean [14]3 years ago

8 0

Answer:

Now e is due to the ring at a

So

We say

1/4πEo(ea/ a²+a²)^3/2

= 1/4πEo ea/2√2a³

So here E is faced towards the ring

Next is E due to a point at the centre

So

E² = 1/4πEo ( e/a²)

Finally we get the total

Et= E²-E

= e/4πEo(2√2-1/2√2)

So the direction here is away from the ring

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Alja [10]

Answer

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3 years ago

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Suppose that in a lightning flash the potential difference between a cloud and the ground is 0.96×109 V and the quantity of char

Dvinal [7]

(a) $2.98\cdot 10^{10} J$

The change in energy of the transferred charge is given by:

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Here we have

$q=31 C$

$\Delta V = 0.96\cdot 10^9 V$

So the change in energy is

$\Delta U = (31 C)(0.96\cdot 10^9 V)=2.98\cdot 10^{10} J$

(b) 7921 m/s

If the energy released is used to accelerate the car from rest, than its final kinetic energy would be

$K=\frac{1}{2}mv^2$

where

m = 950 kg is the mass of the car

v is the final speed of the car

Here the energy given to the car is

$K=2.98\cdot 10^{10} J$

Therefore by re-arranging the equation, we find the final speed of the car:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2.98\cdot 10^{10})}{950}}=7921 m/s$

5 0

3 years ago

Photoelectrons are observed when a metal is illuminated by light with a wavelength less than 383 nm. What is the metal's work fu

Ulleksa [173]

Answer:

Work function of the metal is 3.24 eV.

Explanation:

It is given that,

Wavelength, $\lambda=383\times 10^{-9}\ m$

Let W is the work function of the metal. It is given by using Einstein's photoelectric effect equation. It is given by :

$W=\dfrac{hc}{\lambda}$

$W=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{383\times 10^{-9}}$

$W=5.193\times 10^{-19}\ J$

Since, $1\ eV=1.6\times 10^{-19}\ J$

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3 0

2 years ago

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How long does it take light to reach us from the Sun, 1.53x108 km away? Give your answer in minutes.

muminat

Answer:

Sun light takes 8.5 minutes to reach Earth.

Explanation:

Light travels in vacuum at constant speed $c= 3 \times 10 ^8 \cfrac ms$ , so we can find the time it takes to travel the given distance from Earth to Sun using a kinematic equation.

Finding the time it takes for Sun light to reach us.

Using the kinematic equation $d = vt$ , we can solve for time t, which will give us

$t = \cfrac dv$

We can replace the equation with the given values, but before doing that, we need the values to be on SI units, thus the distance becomes

$d= 1.53 \times 10^8 km \times \cfrac{1000 m}{1 km }\\d = 1.53 \times 10^{11} m$

Replacing the values on the time equation give us

$t = \cfrac{1.53 \times 10^{11} m}{3 \times 10^{8} \cfrac ms}\\t = 510 s$

We get as result 510 seconds, then we must convert them in minutes.

$t = 510 s \times \cfrac{1 min}{60s}\\\boxed{t = 8.5 min }$

So we can conclude that Sun light takes 8.5 minutes to reach Earth.

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