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Two electrons initially repel each other with an electric force of magnitude 2.6×10−20 N .What is the magnitude of the electric field at one electron due to the other?

Respuesta :

boffeemadrid

Answer:

0.1625 N/C

Explanation:

F = Force with which electrons repel each other = [tex]2.6\times 10^{-20}\ N[/tex]

q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

Electric field is given by

[tex]E=\dfrac{F}{q}\\\Rightarrow E=\dfrac{2.6\times 10^{-20}}{1.6\times 10^{-19}}\\\Rightarrow E=0.1625\ N/C[/tex]

The electric field at one electron due to the other is 0.1625 N/C

temdan2001

By using electric field strength formula, the magnitude of the electric field at one electron due to the other is 0.163 N/C

Given that two electrons initially repel each other with an electric force of magnitude 2.6×10^−20 N . That is,

  • Force F = 2.6 x [tex]10^{-20}[/tex] N
  • Charge q = 1.6 x [tex]10^{-19}[/tex] C

The magnitude of the electric field at one electron due to the other can be calculated by using electric field strength formula.

E = F / q

Substitute the two parameters into the formula

E = 2.6 x [tex]10^{-20}[/tex] / 1.6 x [tex]10^{-19}[/tex]

E = 0.1625 N/C

Therefore, the magnitude of the electric field at one electron due to the other is 0.163 N/C approximately.

Learn more about Electric field here: https://brainly.com/question/14372859

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