- #1
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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.