- #1

- 326

- 3

## 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.