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andrey2020 [161]
3 years ago
6

If a proton and an electron are released when they are 4.00 x 10^-10 m apart, find the initial acceleration of each of them.

Physics
2 answers:
SVEN [57.7K]3 years ago
5 0

Answer:

For proton

a=8.8\times 10^{17}\ m/s^2

For electron

a=1.5\times 10^{21}\ m/s^2

Explanation:

We know that

The mass of electron

m=9.1\times 10^{-31}\ kg

The mass of proton

m=1.67\times 10^{-27}\ kg

Charge on electron and proton

q₁=q₂=q

q=1.6\times 10^{-19}\ C

Electrostatics force

F=K\dfrac{q_1q_2}{r^2}

Now by putting the values

F=9\times 10^9\times \dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(4\times 10^{-10})^2}

F=1.44\times 10^{-9}\ N

For proton

F = m a

a =F/m

a=\dfrac{1.4\times 10^{-9}}{1.67\times 10^{-27}}\ m/s^2

a=8.8\times 10^{17}\ m/s^2

For electron

a=\dfrac{1.4\times 10^{-9}}{9.1\times 10^{-31}}\ m/s^2

a=1.5\times 10^{21}\ m/s^2

Morgarella [4.7K]3 years ago
5 0

Answer:

(A) Acceleration of electron a=\frac{1.44\times 10^{-9}}{9.11\times 10^{-31}}=0.158\times 10^{22}m/sec^2

(b) Acceleration of proton a=\frac{1.44\times 10^{-9}}{1.67\times 10^{-27}}=0.8622\times 10^{18}m/sec^2          

Explanation:

We have given distance between proton r=4\times 10^{-10}m

Charge on proton and charge on electron q=1.6\times 10^{-19}

According to coulombs law force between two charge F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_!}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(4\times 10^{-10})^2}=1.44\times 10^{-9}N

Mass of electron m=9.11\times 10^{-31}kg

So acceleration of electron a=\frac{1.44\times 10^{-9}}{9.11\times 10^{-31}}=0.158\times 10^{22}m/sec^2

Mass of proton m=1.67\times 10^{-27}kg

So acceleration of proton  a=\frac{1.44\times 10^{-9}}{1.67\times 10^{-27}}=0.8622\times 10^{18}m/sec^2

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