Question

The electric field near the earth's surface is approximately 150 n/c what is that acceleration

Answer 1

The acceleration of electron and protons are$a = 2.64 \times 10^{13$m/s² $a = 1.44 \times 10^{10$ m/s² respectively.

What is electric field?

Electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects.

Given that,

Magnitude of electric field is

E = 150N/C

a. Acceleration of election?

The charge on an electron is

$q = -1.602\times 10^{-19$

then, the force in electric field is given as

F=qE

$F = 150\times 1.602\times 10^{-19$

$F = 2.403\times10^{-17} N$

Using second law of motion

F= ma

Then, $a =\dfrac{ F}{m}$

Then, mass of electron is

$m = 9.11 \times 10^{-31}kg$

Therefore,

$a =\dfrac{ F}{m}$

$a = \dfrac{2.403\times 10^{-17}}{9.11\times 10^{-31}}$

$a = 2.64 \times 10^{13}$

That’s the magnitude of the electron’s acceleration.

Since the electron has a negative charge the direction of the force on the electron (and also the acceleration) is opposite the direction of the electric field.

b. Same procedure

Acceleration of proton?

The charge on a proton is

$q = +1.602\times 10^{-19}$

then, the force in electric field is given as

F=qE

$F = 150\times 1.602\times 10^{-19}$

$F = 2.403\times 10^{-17} N$

Using second law of motion

F= ma

Then, $a =\dfrac{ F}{m}$

Then, mass of proton is

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

Therefore,

$a =\dfrac{ F}{m}$

$a = \dfrac{2.403\times 10{^-17}}{1.67\times 10^{-27}}$

a = 1.44 ×10^10m/s²

That’s the magnitude of the proton's acceleration.

Since the proton has a positive charge the direction of the force on the proton (and also the acceleration) is in the same direction of the electric field.

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