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An electron and a proton are fixed at a separation distance of 973 nm. Find the magnitude and the direction of the electric field at their midpoint.

Respuesta :

whitneytr12

Answer:

The magnitude is: [tex]|E|=6084.1\: N/C[/tex]

The direction of E is in the negative x-direction.

Explanation:

The electric field equation is:

[tex]E=k\frac{Q}{r^{2}}[/tex]

Where:

  • Q is the charge (we can choose the electron or the proton)
  • r is the distance (in our case is at the midpoint 973/2 nm)
  • k is the Coulomb constant ([tex]9*10^{9}\: Nm^{2}C^{-2}[/tex])

Using the electron charge ([tex]e = -1.6*10^{-19}\: C[/tex])

[tex]E=-9*10^{9}\frac{1.6*10^{-19}}{(486.5*10^{-9})^{2}}[/tex]

The magnitude is:

[tex]|E|=6084.1\: N/C[/tex]

The direction of E is in the negative x-direction.

I hope it helps you!

boffeemadrid

The magnitude and direction of an electric field is required.

The electric field at the midpoint is [tex]12168.21\ \text{N/m}[/tex] the direction is towards electron.

k = Coulomb constant = [tex]9\times 10^9\ \text{Nm}^2/\text{C}^2[/tex]

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

r = Distance between the electron and proton = 973 nm

The electric field will be

[tex]E=\dfrac{2kq}{\left(\dfrac{1}{2}r\right)^2}\\\Rightarrow E=\dfrac{8kq}{r^2}\\\Rightarrow E=\dfrac{8\times 9\times 10^9\times 1.6\times 10^{-19}}{(973\times 10^{-9})^2}\\\Rightarrow E=12168.21\ \text{N/m}[/tex]

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