Calculating Electron Force in a Two Point Charge System - Physics Homework Help

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The problem involves calculating the force on an electron placed halfway between two point charges, one positive and one negative, each with a magnitude of 1.62 μC and separated by 47.8 cm. The electric field at the electron's position is derived using the formula E = 2kQ/(d/2)², leading to a calculated force of 8.17 x 10^-14 N. The direction of the force on the electron is toward the positive charge, as electrons are attracted to positive charges. The calculations and reasoning presented for both the force magnitude and direction are confirmed to be correct. This analysis effectively demonstrates the application of Coulomb's law in a two-point charge system.
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Homework Statement



Two point charges are distance d = 47.8 cm apart, They have equal and opposite charge, each with magnitude Q = 1.62 μC. The negative charge is on the left, and the positive charge is on the right. An electron (mass ##9.11\times 10^{-31}## kg, charge ##1.6\times 10^{-19}## C) is placed half-way between the charges.

c) What is the force of the electron?

d) What is the direction of the force on the electron?

Homework Equations



  • ##E = \dfrac{kQ}{r^2}##
  • ##F = Eq##

The Attempt at a Solution



I just want to make sure if I'm right for both parts of the problem

For part (c), I got this

##E = \dfrac{2kQ}{(\frac{d}{2})^2}##

##F = Eq##
##F = \dfrac{2kQq}{(\frac{d}{2})^2}##

where

##d = \text{distance between the two charges before the electron is placed}##
##q = \text{charge of the electron}##
##Q = \text{point charge with given value}##

By substitution, I got ##8.17 \times 10^{-14}## N.

For part (d), I believe the answer is "right, toward the positive charge" since electron attracts with the point charge with positive sign.
 
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Looks good to me: I didn't check your calculator skill, but the physics seems right.
 
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