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3 years ago

14

Why the center of Newton rings is dark?

2 answers:

RideAnS [48]3 years ago

3 0

The point of contact the path difference is zero but one of the interfering ray is reflected so the effective path difference becomes λ/2 thus the condition of minimum intensity is created in the center.

grandymaker [24]3 years ago

3 0

Answer:

the point of contact Thè path different is zero but

<h3>one of inter ferying ray is reflected so the effective path difference become<u> </u><u>/</u> 2 and in this condition minimum intensity is created at center so the center of Newton ring is dark</h3>

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Fat Albert (the TV show character) runs up the stairs on Monday. On Tuesday, he walks up the same set of stairs. Which day did h

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A force of 2 kN is applied to an object to make it move 3.6 m in the direction of the force. Select the correct value of work do

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3 years ago

Un prisma de cemento pesa 2500 N y ejerce una presión de 125 Pa, ¿cuál es el valor del área en la cual se apoya?

Eduardwww [97]

Answer:

Area = 20 m²

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3 years ago

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3 years ago

Read 2 more answers

A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se

zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, $\lambda = 1\ nC = 1\times 10^{- 9}\ C$

Charge density of rod 2, $\lambda = - 1\ nC = - 1\times 10^{- 9}\ C$

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

$\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}$

Also,

$\vec{E} = \frac{2K\lambda }{R}$ (1)

where

K = electrostatic constant = $\frac{1}{4\pi \epsilon_{o} R}$

R = Distance

$\lambda$ = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

$\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C$

$\vec{E} = 1800\ N/C$ (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

$\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C$

$\vec{E'} = 1800\ N/C$ (towards)

Now, the total field at the origin is the sum of both the fields:

$\vec{E_{net}} = 1800 + 1800 = 3600\ N/C$

7 0

3 years ago