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Sauron [17]
3 years ago
7

Two imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the

concentric spheres. When compared to the number of field lines N1 going through the sphere of radius R, the number of electric field lines N2 going through the sphere of radius 2R is
Physics
1 answer:
Andrej [43]3 years ago
4 0

Answer:

N2 = ¼N1

Explanation:

First of all, let's define the terms;

N1 = number of electric field lines going through the sphere of radius R

N2 = number of electric field lines going through the sphere of radius 2R

Q = the charge enclosed at the centre of concentric spheres

ε_o = a constant known as "permittivity of the free space"

E1 = Electric field in the sphere of radius R.

E2 = Electric field in the sphere of radius 2R.

A1 = Area of sphere of radius R.

A2 = Area of sphere of radius 2R

Now, from Gauss's law, the electric flux through the sphere of radius R is given by;

Φ = Q/ε_o

We also know that;

Φ = EA

Thus;

E1 × A1 = Q/ε_o

E1 = Q/(ε_o × A1)

Where A1 = 4πR²

E1 = Q/(ε_o × 4πR²)

Similarly, for the sphere of radius 2R,we have;

E2 = Q/(ε_o × 4π(2R)²)

Factorizing out to get;

E2 = ¼Q/(ε_o × 4πR²)

Comparing E2 with E1, we arrive at;

E2 = ¼E1

Now, due to the number of lines is proportional to the electric field in the each spheres, we can now write;

N2 = ¼N1

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Answer:

649kg/m^3

Explanation:

Let p be the density of this particular object.

Formula for density:

p =  \frac{mass \: (in \: kg)}{volume \: (in \:  {m}^{3}) }

We can substitute the givenmass and volume to find density of the object.

p =  \frac{25kg}{0.0385 {m}^{3} }  \\  = 649kg \: per \:  {m}^{3}

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current is passed through two parallel conductors in the same direction. If the conductors are placed near each other,they will​
svlad2 [7]

If current is passed through two parallel conductors in the same direction and the conductors are placed near each other, they will​ attract each other.

<h3>What is electric current?</h3>

Electric current can be defined as the flow of electrons.

Since electrons are easily removed from atom and are very mobile, the flow of electrons constitute an electric current.

Materials which allow electric current to flow through them are known as conductors. Examples of conductors are metals, and electrolytes.

On the other hand, materials which do not allow electric current to pass through them are known as insulators. Examples of insulators are wood and rubber.

The flow of current is known as electricity.

Parallel conductors with current flowing through them in the same direction are attracted to each other as a result of a magnetic field produced by the flow of current.

In conclusion, conductors allow electric current to pass through and the flow of current through a conductor produces a magnetic field.

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An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s
max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

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You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

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