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antoniya [11.8K]
3 years ago
11

A voltage of 18 V is applied across the ends of a piece of copper wire 8 cm long. The mass of an electron is 9.11 x 10kg and its

charge is 1.6 x 10-B c What is the magnitude of the electron's acceleration? Answer in units of m/s part 2 of 2 points After traveling 4 × 10-8 m, if it does not collide with a copper ion over this distance what is the kinetic energy of the electron
Physics
1 answer:
saw5 [17]3 years ago
7 0

Explanation:

It is given that,

Voltage, V = 18 V

Length of the copper wire, l = 8 cm = 0.08 m

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

Charge on electron, q=1.6\times 10^{-19}\ C

Electric force is given by, F=qE=\dfrac{qV}{l}

Also, ma=\dfrac{qV}{l}

a=\dfrac{qV}{lm}

a=\dfrac{1.6\times 10^{-19}\times 18}{0.08\times 9.11\times 10^{-31}}

a=3.95\times 10^{13}\ m/s^2

Since, the electron is at rest initially, u = 0

v^2=2as, s=4\times 10^{-8}\ m

v^2=2\times 3.95\times 10^{13}\times 4\times 10^{-8}

v^2=3160000

The kinetic energy of the electron is :

E=\dfrac{1}{2}\times m\times v^2

E=\dfrac{1}{2}\times 9.11\times 10^{-31}\times (3160000)^2

E=4.54\times 10^{-18}\ J

Hence, this is the required solution.

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