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Salsk061 [2.6K]
4 years ago
9

Calculate the potential V(r) for r>rb. (Hint: The net potential is the sum of the potentials due to the individual spheres.)

Use ϵ0 as the permittivity of free space and express your answer in terms of some or all of the variables r, ra, rb, q, and any appropriate constants.
Physics
1 answer:
Marat540 [252]4 years ago
8 0

Answer:

The potential for r > rb is equal to zero.

Explanation:

For r > rb, the potential is:

V=\frac{Kq}{r}

Then, the net potential is:

V_{(r)} =\frac{K(+\epsilon )}{r} +\frac{K(-\epsilon )}{r}

K=\frac{1}{4\pi \epsilon _{o}  }

V_{(r)} =\frac{K(+\epsilon )}{r} -\frac{K(\epsilon )}{r}\\V_{(r)}=0

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Read 2 more answers
As it travels through a crystal, a light wave is described by the function E(x,t)=Acos[(1.57×107)x−(2.93×1015)t]. In this expres
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Answer:

Speed, v=1.86\times 10^8\ m/s

Explanation:

It is given that,

A light wave is described by the following function as :

E(x,t)=A\ cos[(1.57\times 10^7)x-(2.93\times 10^{15})t].....(1)

The general equation of wave is given by :

E=Acos(kx-\omega t)........(2)

On comparing equation (1) and (2)

k=(1.57\times 10^7)

\dfrac{2\pi}{\lambda}=(1.57\times 10^7)

\lambda=\dfrac{2\pi}{(1.57\times 10^7)}

Wavelength, \lambda=4.002\times 10^{-7}\ m

\omega=(2.93\times 10^{15})

\dfrac{2\pi}{T}=(2.93\times 10^{15})

\dfrac{1}{T}=\dfrac{(2.93\times 10^{15})}{2\pi}

Frequency, f=4.66\times 10^{14}\ Hz

Let v is the speed of the light wave. It is given by :

v=f\times \lambda

v=4.66\times 10^{14}\ Hz\times 4.002\times 10^{-7}\ m

v=1.86\times 10^8\ m/s

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