# Question b8f2a

##### 2 Answers
Apr 19, 2016

If the charges are separated by a distance $r = 10 m$, then each charge would need a magnitude $q = 2.33 \times {10}^{-} 3 C$ in order for the force between them to equal the weight of a 50 kg person.

#### Explanation:

We'll start with the 6-step solution below and then follow-up with a more detailed explanation.

1) Let ${F}_{W} = {F}_{C}$

2) ${F}_{C} = \frac{1}{4 \cdot \pi \cdot {\epsilon}_{0}} \cdot \frac{{q}_{1} \cdot {q}_{2}}{{r}^{2}}$

3) ${F}_{W} = m \cdot g$

4) So re-write ${F}_{W} = {F}_{C}$ as

${F}_{W} = m \cdot g = \frac{1}{4 \cdot \pi \cdot {\epsilon}_{0}} \cdot \frac{{q}_{1} \cdot {q}_{2}}{{r}^{2}} = {F}_{C}$

5) $\left({q}_{1} \cdot {q}_{2}\right) = m \cdot g \cdot \left(4 \cdot \pi \cdot {\epsilon}_{0}\right) \cdot \left({r}^{2}\right)$

6) Since ${q}_{1} = {q}_{2}$, we can write ${q}_{1}^{2} = m \cdot g \cdot \left(4 \cdot \pi \cdot {\epsilon}_{0}\right) \cdot \left({r}^{2}\right)$

${q}_{1}^{2} = 50 k g \cdot 9.81 \frac{m}{s} ^ 2 \cdot \left(4 \cdot 3.1415 \cdot 8.854 \cdot {10}^{-} 12 \frac{F}{m}\right) \cdot {\left(10 m\right)}^{2}$
${q}_{1}^{2} = \sqrt{5.45 \times {10}^{-} 6 {C}^{2}}$
${q}_{1} = 2.33 \times {10}^{-} 3 C$

...and we’re done!

1) The problem states that “the force between [the charges] equals the weight of a 50 kg person”. ${F}_{C} = {F}_{W}$ is a simple, way to express this mathematically, where ${F}_{C}$ represents the force between the charges and ${F}_{W}$ is the weight of the person.

2) ${F}_{C}$ is the (electrostatic) force acting between two charges separated by a distance $r$, where the magnitude of one of the charges is, ${q}_{1}$ and the magnitude of the other charge is ${q}_{2}$. ${\epsilon}_{0}$ is the electric permeability of free space.

3) ${F}_{W} = m \cdot g$ tells us that the weight of the person [or any object for that matter] is the mass of the object, $m$, times the acceleration of gravity [due to the earth’s gravitational pull].

4) Here we expressed 1) in more explicit form, using equations 2) and 3)

5) The problem asks “What should be the magnitude of the charges”, so we rearrange equation 4) to solve for the charges, ${q}_{1} \cdot {q}_{2}$, by multiplying both sides by $4 \cdot \pi \cdot {\epsilon}_{0} \cdot {r}^{2}$.

6) The phrase “Two Equal charges” indicates that we can set ${q}_{2} = {q}_{1}$. In other words, the magnitude of each charge is the same. So ${q}_{1} \cdot {q}_{2} = {q}_{1}^{2}$. After that, it’s just a matter of substituting the values for the quantities ($m , g , \pi , {\epsilon}_{0} \mathmr{and} r$ ), carrying out a little arithmetic and we’re done!

Apr 21, 2016

$2.34 \times {10}^{-} 3 C$, rounded to two decimal places.

#### Explanation:

Force between two charges ${q}_{1} \mathmr{and} {q}_{2}$ at a distance $r$ is given by Coulomb's Law

${F}_{C} = {k}_{e} | {q}_{1} {q}_{2} \frac{|}{r} ^ 2$,

where ${k}_{e}$ is Coulombs' Constant and =8.99xx10^9 N m^2 C^(−2)#
Force ${F}_{g}$ due to weight of a person$= m g$, where $m$ is person's mass and $g$ acceleration due to gravity and has value of $9.81 m {s}^{-} 2$.

Since given charges are equal, let each be $= q$
Equating both forces and inserting given values we obtain
$8.99 \times {10}^{9} | {q}^{2} \frac{|}{10} ^ 2 = 50 \times 9.81$, simplifying and solving for $q$
$| q | = \sqrt{\frac{50 \times 9.81}{8.99 \times {10}^{7}}}$
$| q | = 2.34 \times {10}^{-} 3 C$, rounded to two decimal places.