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3 years ago

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Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 570 N/C. If the particle

s are free to move, what are their speeds (in m/s) after 47.6 ns?

1 answer:

Veseljchak [2.6K]3 years ago

3 0

Explanation:

Given that,

Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 570 N/C.

We need to find their speeds after 47.6 ns.

For electron,

The electric force is given by :

$F=qE\\\\F=1.6\times 10^{-19}\times 570\\\\=9.12\times 10^{-17}\ N$

Let a be the acceleration of the electron. So,

F = ma

m is mass of electron

$a=\dfrac{F}{m}\\\\a=\dfrac{9.12\times 10^{-17}}{9.1\times 10^{-31}}\\\\a=10^{14}\ m/s^2$

Let v be the final velocity of the electron. So,

v = u +at

u = 0 (at rest)

So,

$v=10^{14}\times 47.6\times 10^{-9}\\\\v=4.76\times 10^6\ m/s$

For proton,

Acceleration,

$a=\dfrac{9.12\times 10^{-17}}{1.67\times 10^{-27}}\\\\=5.46\times 10^{10}\ m/s^2$

Now final velocity of the proton is given by :

$v=5.46\times 10^{10}\times 47.6\times 10^{-9}\\\\v=2598.96\ m/s$

Hence, this is the required solution.

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Explanation:

1)

The electric field inside the sphere is given by

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The potential at the surface, V(R), is that of a point charge, so

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Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

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On the other hand, the potential at the surface is

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The graph of V versus r can be found in attachment.

We observe the following:

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- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

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