Question

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.

Answer 1

Answer:

The magnitude is: $|E|=6084.1\: N/C$

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

Explanation:

The electric field equation is:

$E=k\frac{Q}{r^{2}}$

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 ($9*10^{9}\: Nm^{2}C^{-2}$)

Using the electron charge ($e = -1.6*10^{-19}\: C$)

$E=-9*10^{9}\frac{1.6*10^{-19}}{(486.5*10^{-9})^{2}}$

The magnitude is:

$|E|=6084.1\: N/C$

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

I hope it helps you!