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soldi70 [24.7K]
3 years ago
14

A very long cylinder, of radius a, carries a uniform polarization P perpendicular to its axis. Find the electric field inside th

e cylinder (in terms of P). Show that the field outside the cylinder can be expressed (in cylindrical coordinates) in the form

Physics
2 answers:
Zielflug [23.3K]3 years ago
6 0

Answer:     Eout= (1/2\epsilon)(R/r)^2(2((P . r )r-P)

Explanation:

The requires extensive derivation. The equation has been attached and answer is as follows.

Let the positively charged end as +d/2 and negatively charged end to be       -d/2.

Look for the Gauss's law in images attached which is E= (ρ/2ε)r

The formulas in the image below provides the formulas for the cylinders which are offset.

The total field inside the cylinder is now equals to adding the two is now  equals to             Ein= -P/2ε

In the second picture of the sketch attached it shows that the field outside is one of two lines of charge with linear densities of λ and offset by d:

Means that, λ=ρR² π

Continue along image no 2 and 3 attached to figure out the correct identities used in order to get to the required solution.

Remember the first images are the drawn sketches in order to imagine the electric field along the surface ( which here is the cylinder ).

Juliette [100K]3 years ago
4 0

Answer:

For electric field inside cylinder, check image 02 attached

For electric field outside cylinder, check image 03 attached

Explanation:

Let's consider the polarized cylinder as superposition of two cylinders with opposite,equal, uniform charge densities in a way shown in the figure in the "image 01"d attached ;

In general, if we have an object with polarization (P¬) , then we have to take two objects with similar shape to the system, with opposite, equal, and uniform charge densities and then we super-impose these two objects in such a way that the total dipole moment of this superimposed system is equal to the total dipole moment of original system.

Now, we can take the super- imposed system as equivalent to the original system for calculating electric field and potential.

Therefore,

For the electric field inside the cylinder, check the solution in "image 02" i attached

For the electric field outside the cylinder, check, "image 03" i attached.

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4 0
2 years ago
Two resistors, of R1 = 3.93 Ω and R2 = 5.59 Ω, are connected in series to a battery with an EMF of 24.0 V and negligible interna
son4ous [18]

Answer:

2.521 (A); 14.0924 (V)

Explanation:

more info in the attachment, the answers are marked with red colour.

4 0
2 years ago
Which Box in the diagram above, should this picture go into?
lyudmila [28]
What diagram? There isn’t one
7 0
3 years ago
What's 600,000,000 divided by 3,000.000,000,000?
Murljashka [212]

Answer:0.000002

Explanation: I Looked It Up lol

5 0
3 years ago
THIS MARCIN
nekit [7.7K]

Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

f

1

=

v

1

−

u

1

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

h

h

′

=

u

v

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

h

h

′

=

u

v

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

f

1

=

v

1

−

(−25)

1

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

v

1

=

f

1

+

(−25)

1

=

10

1

−

25

1

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

v

1

=

250

25−10

=

250

15

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

15

250

=

3

50

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

5

h

′

=−

25

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

25

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

5 0
2 years ago
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