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3 years ago

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A rod of length L has a charge Q uniformly distributed along its length. The rod lies along the y axis with one end at the origi

n. (a) Find an expression for the electric potential as a function of position along the x axis.

1 answer:

umka21 [38]3 years ago

3 0

Answer:

V = k Q/l ln [(l +√(l² + x²)) / x]

Explanation:

The electrical potential for a continuous distribution of charges is

V = k ∫ dq / r

Let's apply this expression to our case, define a linear charge density for the bar

λ = dq / dy

dq = λ dy

The distance from a point on the bar to the x-axis is

r = √ (x² + y²)

Let's replace

V = K ∫ λ dy /√ (x² + y²)

We integrate

V = k λ ln (y + √ (x² + y²))

Let's evaluate between y = 0 and y = l

V = k λ [ ln (l +√(x² + l²) - ln x]

We substitute the linear density

V = k Q/l ln [(l +√(l² + x²)) / x]

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Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid

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Answer:

a

$z= 2.5 \ m$

b

$z = (1 \ m , 4 \ m )$

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From the question we are told that

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The wavelength of each source wave $\lambda = 6.0 \ m$

Let the distance from source A where the construct interference occurred be z

Generally the path difference for constructive interference is

$z - (d-z) = m \lambda$

Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0

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Generally the path difference for destructive interference is

$|z-(d-z)| = (2m + 1)\frac{\lambda}{2}$

=> $|2z - d |= (0 + 1)\frac{\lambda}{2}$

=> $|2z - d| =\frac{\lambda}{2}$

substituting values

$|2z - 5| =\frac{6}{2}$

=> $z = \frac{5 \pm 3}{2}$

So

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MArishka [77]

Answer:

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$I_c = \frac{I}{I_o} =8.48 *10^{-8}$

Explanation:

From the question we are told that

The width of the slit is $D = 0.3 \ mm = 0.3 *10^{-3} \ m$

The wavelength is $\lambda = 254 \ nm = 254 *10^{-9} \ m$

The angle is $\theta = 11^o$

The intensity of at $11^o$ to the axis in terms of the intensity of the central maximum. is mathematically represented as

$I_c = \frac{I}{I_o} = [ \frac{sin \beta }{\beta }] ^2$

Where $\beta$ is mathematically represented as

$\beta = \frac{D sin (\theta ) * \pi}{\lambda }$

substituting values

$\beta = \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }$

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So

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