Answer:
ᵃ. Fₙ= 1.424 x 10¹³ μN
ᵇ. away from the +6 charge
Explanation:
PART A
Using Coulomb's law FOR CHARGES +2 AND +4
F= (Q . q)÷4(3.142)ε(5/100)²
F= 2.848 x 10¹³ μN
for charges of 2 and 6
F₂=(Q . q)÷4(3.142)ε(5/100)
F₂= 4.272 x 10¹³ μN
<em>NET FORCE IS EQUAL TO THE SUM OF ALL THE FORCES ACTING AT A POINT.</em>
<em>FORCE IS A VECTOR QUANTITY, THEREFORE, WE ASSIGN A NEGATIVE SIGN TO ONE OF THE FORCES, as both of them are acting in opposite directions while repelling the +2. </em>
Fₙ= F-F₂
Fₙ= 4.272 x 10¹³ - 2.848 x 10¹³
Fₙ= 1.424 x 10¹³ μN
PART B
The force caused by the +6 charge is twice as much as the one caused by the +4 charge on the same +2 charge, therefore the repulsion caused by the greater charge is greater and has an effect of pushing the +2 charge away from it. However there is also a force of repulsion by the +4 charge which is NOT AS STRONG as compared to the +6 charge, therefore it is less likely to push away the +2 charge
