Question

A test charge of +2 μC is placed halfway between a charge of +6 μC and another of +4 μC separated by 10 cm. (a) What is the magnitude of the force on the test charge? (b) What is the direction of this force (away from or toward the +6 μC charge)?

Answer 1

ᵃ. Fₙ= 1.424 x 10¹³ μN

ᵇ. away from the +6 charge

Step-by-step 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

NET FORCE IS EQUAL TO THE SUM OF ALL THE FORCES ACTING AT A POINT.

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.

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