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Hi can someone help me please ,Let d= 1 for second part

Hi Can Someone Help Me Please Let D 1 For Second Part class=

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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