Help me?!?!?!?!?!?!?!?!?!?!?!?!?!

Physics

2 answers:

fredd [130]4 years ago

8 0

Before you can pick the correct one, you have to review in your mind what
you know about the electric field. (Or what you SHOULD know about it.)

-- The field lines show where a teeny tiny POSITIVE charge
wants to go when you put it down anywhere.

-- A teeny tiny POSITIVE charge always wants to get AWAY from other
positive charges, and it wants to go TOWARD negative charges.

That's all you need. Now let's look at the pictures.
I want to look at them from the bottom up:

(D). ... No.
The arrows show that the tiny positive test charge wants to go AWAY from
the negative (blue) charge and TOWARD the positive (red) charge.
That's wrong.

(C). ... No.
The arrows show that the tiny positive test charge wants to go AWAY from
any negative charges. That's wrong.

(B). ... No.
The arrows show that the tiny positive test charge wants to go AWAY from
BOTH the negative charge AND the positive charge. That's wrong.

(A). ... YES !
The arrows show that the tiny test charge wants to go away from the positive
charge, and toward the negative charge. That's exactly right.

(That's why I wanted to do them backwards ... because
the first picture on the page is the correct one.)

marishachu [46]4 years ago

3 0

The correct field line would be A.

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Monochromatic light of wavelength λ=620nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is

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Answer:

The intensity of light from the 1mm from the central maximu is $I = 0.822I_o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 620 nm = 620 *10^{-9}m$

The width of the slit is $w = 0.450mm = \frac{0.45}{1000} = 0.45*10^{-3} m$

The distance from the screen is $D = 3.00m$

The intensity at the central maximum is $I_o$

The distance from the central maximum is $d_1 = 1.00mm = \frac{1}{1000} = 1.0*10^{-3}m$

Let z be the the distance of a point with intensity I from central maximum

Then we can represent this intensity as

$I = I_o [\frac{sin [\frac{\pi * w * sin (\theta )}{\lambda} ]}{\frac{\pi * w * sin (\theta )}{\lambda } } ]^2$

Now the relationship between D and z can be represented using the SOHCAHTOA rule i.e

$sin \theta = \frac{z}{D}$

if the angle between the the light at z and the central maximum is small

Then $sin \theta = \theta$

Which implies that

$\theta = \frac{z}{D}$

substituting this into the equation for the intensity

$I = I_o [\frac{sin [\frac{\pi w}{\lambda} \cdot \frac{z}{D} ]}{\frac{\pi w z}{\lambda D\frac{x}{y} } } ]$

given that $z =1mm = 1*10^{-3}m$

We have that

$I = I_o [\frac{sin[\frac{3.142 * 0.45*10^{-3}}{(620 *10^{-9})} \cdot \frac{1*10^{-3}}{3} ]}{\frac{3.142 * 0.45*10^{-3}*1*10^{-3} }{620*10^{-9} *3} } ]^2$