An electron a distance $d$ from a proton feels an electrostatic force of magnitude $\left|\vec{F}_{e}\right|$. Imagine that we move the electron so that it is a distance $2 d$ from a helium nucleus (which has two protons and two neutrons). The magnitude of the force on the electron now is $b\left|\vec{F}_{e}\right|$, where $b$ is
A) 4
B) 2
C) 1
D) $\mathbf{1 / 2}$
E) $1 / 4$

Question

Quizplus · Accepted Answer

The force between charged particles is given by Coulomb's law, $F = k \frac{|q_1 q_2|}{r^2}$, where $k$ is Coulomb's constant, $q_1$ and $q_2$ are the charges, and $r$ is the distance between the charges. When the electron is moved to a distance $2d$ from a helium nucleus, the distance $r$ becomes $2d$, and the charge of the helium nucleus ($q_2$) is twice that of a proton due to having two protons. Therefore, the force becomes $F = k \frac{2|q_e q_p|}{(2d)^2}$, which simplifies to $F = \frac{1}{2} k \frac{|q_e q_p|}{d^2}$, or $\frac{1}{2}|\vec{F}_{e}|$. Thus, $b = \frac{1}{2}$.

An Electron a Distance ddd From a Proton Feels an Electrostatic Force of Magnitude

An Electron a Distance $d$ From a Proton Feels an Electrostatic Force of Magnitude