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

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SUMMARY

The discussion focuses on calculating the force experienced by an electron placed between two point charges, each with a magnitude of Q = 1.62 μC and separated by a distance of d = 47.8 cm. The user correctly applies the formula for electric field strength, E = kQ/r², and subsequently calculates the force using F = Eq, resulting in a force of 8.17 × 10⁻¹⁴ N. The direction of the force is confirmed to be toward the positive charge, as the electron is attracted to it due to opposite charges.

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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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