Question

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

Answer 1

`\begin{gathered} \text{blank}1\colon(1)/(4)F_e \\ \text{.} \\ \text{.} \end{gathered}`

2)

`F=2.3097\cdot10^(48)\text{ Newtons}`

Step-by-step explanation

the electrostatic force between two forces is given by:

`\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}`

then, complete the table we need to use this equation

where

`F=k\frac{q_1q_2}{d^2_{}}`

k, q1 and q2 are ginven in the table, so

Step 1

Let

`\begin{gathered} charge_1=(1)/(4) \\ charge_2=1 \\ \text{distance}=1 \end{gathered}`

so,replace

`\begin{gathered} F=k\frac{q_1q_2}{d_{}} \\ F=1\frac{(1)/(4)\cdot1}{1_{}}=(1)/(4) \end{gathered}`

hence, for tha row

`\begin{gathered} (1)/(4)F_e \\ \end{gathered}`

Step 2

now, do the same for the next Fe, so

b)Let

`\begin{gathered} charge_1=1 \\ charge_2=(1)/(2) \\ \text{distance}=1 \end{gathered}`

so,replace

`\begin{gathered} F=k\frac{q_1q_2}{d_{}} \\ F=1\frac{1\cdot(1)/(2)}{1_{}}=(1)/(2) \end{gathered}`

hence, for tha row

`\begin{gathered} (1)/(2)F_e \\ \end{gathered}`

Step 3

now, do the same for the next Fe, so

c)Let

`\begin{gathered} charge_1=1 \\ charge_2=(1)/(4) \\ \text{distance}=1 \end{gathered}`

so,replace

`\begin{gathered} F=k\frac{q_1q_2}{d_{}} \\ F=1\frac{1\cdot(1)/(4)}{1_{}}=(1)/(4) \end{gathered}`

hence, for that row

`\begin{gathered} (1)/(4)F_e \\ \end{gathered}`

Step 4

now, do the same for the next Fe, so

c)Let

`\begin{gathered} charge_1=(1)/(2) \\ charge_2=(1)/(2) \\ \text{distance}=1 \end{gathered}`

so,replace

`\begin{gathered} F=k\frac{q_1q_2}{d_{}} \\ F=1\frac{(1)/(2)\cdot(1)/(2)}{1_{}}=(1)/(4) \end{gathered}`

hence, for tha row

`\begin{gathered} (1)/(4)F_e \\ \end{gathered}`

Step 5

now, do the same for the next Fe, so

c)Let

`\begin{gathered} charge_1=(1)/(8) \\ charge_2=(1)/(4) \\ \text{distance}=1 \end{gathered}`

so,replace

`\begin{gathered} F=k\frac{q_1q_2}{d_{}} \\ F=1\frac{(1)/(8)\cdot(1)/(4)}{1_{}} \\ F=1\frac{(1)/(32)}{1_{}}=(1)/(32) \end{gathered}`

hence, for that row

`\begin{gathered} (1)/(32)F_e \\ \end{gathered}`

Step 6

find the electrostastic force betweeen electrons

let

`\begin{gathered} q_1=q_2=-\text{1}.602\cdot10^(19)\text{ C} \\ k=9\cdot10^9(Nm^2)/(C^2) \end{gathered}`

as the distance is not give, let's use

distance =1

replace in the formula

`\begin{gathered} F=k\frac{q_1q_2}{d_{}} \\ F=9\cdot10^9(Nm^2)/(C^2)\frac{(^2-\text{1}.602\cdot10^(19)C)}{1_{}} \\ F=2.3097\cdot10^(48)\text{ N} \end{gathered}`

therefore, the electrostaic force is

`F=2.3097\cdot10^(48)\text{ Newtons}`

I hope this helps you