Sagot :
[tex]\begin{gathered} \text{blank}1\colon\frac{1}{4}F_e \\ \text{.} \\ \text{.} \end{gathered}[/tex]
2)
[tex]F=2.3097\cdot10^{48}\text{ Newtons}[/tex]Explanation
the electrostatic force between two forces is given by:
[tex]\begin{gathered} F=k\frac{q_1q_2}{d^2_{}} \\ where\text{ } \\ k\text{ is a constant}(in\text{ this case we n}eed\text{ asume k=1)} \\ q_1\text{ is the charge 1} \\ q_2\text{ is the charge }2 \\ \text{and d is the distance betw}en\text{ them} \end{gathered}[/tex]then, complete the table we need to use this equation
where
[tex]F=k\frac{q_1q_2}{d^2_{}}[/tex]k, q1 and q2 are ginven in the table, so
Step 1
Let
[tex]\begin{gathered} charge_1=\frac{1}{4} \\ charge_2=1 \\ \text{distance}=1 \end{gathered}[/tex]so,replace
[tex]\begin{gathered} F=k\frac{q_1q_2}{d_{}} \\ F=1\frac{\frac{1}{4}\cdot1}{1_{}}=\frac{1}{4} \end{gathered}[/tex]hence, for tha row
[tex]\begin{gathered} \frac{1}{4}F_e \\ \end{gathered}[/tex]Step 2
now, do the same for the next Fe, so
b)Let
[tex]\begin{gathered} charge_1=1 \\ charge_2=\frac{1}{2} \\ \text{distance}=1 \end{gathered}[/tex]so,replace
[tex]\begin{gathered} F=k\frac{q_1q_2}{d_{}} \\ F=1\frac{1\cdot\frac{1}{2}}{1_{}}=\frac{1}{2} \end{gathered}[/tex]hence, for tha row
[tex]\begin{gathered} \frac{1}{2}F_e \\ \end{gathered}[/tex]Step 3
now, do the same for the next Fe, so
c)Let
[tex]\begin{gathered} charge_1=1 \\ charge_2=\frac{1}{4} \\ \text{distance}=1 \end{gathered}[/tex]so,replace
[tex]\begin{gathered} F=k\frac{q_1q_2}{d_{}} \\ F=1\frac{1\cdot\frac{1}{4}}{1_{}}=\frac{1}{4} \end{gathered}[/tex]hence, for that row
[tex]\begin{gathered} \frac{1}{4}F_e \\ \end{gathered}[/tex]Step 4
now, do the same for the next Fe, so
c)Let
[tex]\begin{gathered} charge_1=\frac{1}{2} \\ charge_2=\frac{1}{2} \\ \text{distance}=1 \end{gathered}[/tex]so,replace
[tex]\begin{gathered} F=k\frac{q_1q_2}{d_{}} \\ F=1\frac{\frac{1}{2}\cdot\frac{1}{2}}{1_{}}=\frac{1}{4} \end{gathered}[/tex]hence, for tha row
[tex]\begin{gathered} \frac{1}{4}F_e \\ \end{gathered}[/tex]Step 5
now, do the same for the next Fe, so
c)Let
[tex]\begin{gathered} charge_1=\frac{1}{8} \\ charge_2=\frac{1}{4} \\ \text{distance}=1 \end{gathered}[/tex]so,replace
[tex]\begin{gathered} F=k\frac{q_1q_2}{d_{}} \\ F=1\frac{\frac{1}{8}\cdot\frac{1}{4}}{1_{}} \\ F=1\frac{\frac{1}{32}}{1_{}}=\frac{1}{32} \end{gathered}[/tex]hence, for that row
[tex]\begin{gathered} \frac{1}{32}F_e \\ \end{gathered}[/tex]Step 6
find the electrostastic force betweeen electrons
let
[tex]\begin{gathered} q_1=q_2=-\text{1}.602\cdot10^{19}\text{ C} \\ k=9\cdot10^9\frac{Nm^2}{C^2} \end{gathered}[/tex]as the distance is not give, let's use
distance =1
replace in the formula
[tex]\begin{gathered} F=k\frac{q_1q_2}{d_{}} \\ F=9\cdot10^9\frac{Nm^2}{C^2}\frac{(^2-\text{1}.602\cdot10^{19}C)}{1_{}} \\ F=2.3097\cdot10^{48}\text{ N} \end{gathered}[/tex]therefore, the electrostaic force is
[tex]F=2.3097\cdot10^{48}\text{ Newtons}[/tex]I hope this helps you
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