Anegative charge, q1, of 6 µc is 0.002 m north of a positive charge, q2, of 3 µc. what is the magnitude and direction of the electrical force, fe, applied by q1 on q2? a) magnitude: 8 × 10^1 n direction: south b) magnitude: 8 × 10^1 n direction: north c) magnitude: 4 × 10^4 n direction: south d) magnitude: 4 × 10^4 n direction: north

D. magnitude: 4 × 104 N

direction: north

D) magnitude: 4 × 10^4 N

direction: north

Explanation:

q₁ = magnitude of negative charge = 6 x 10⁻⁶ C

q₂ = magnitude of positive charge = 3 x 10⁻⁶ C

r = distance between the two charges = 0.002 m

Magnitude of electric force between the two charges is given as

F = k q₁ q₂/r²

where k = constant = 9 x 10⁹

inserting the values

F = (9 x 10⁹) (6 x 10⁻⁶) (3 x 10⁻⁶)/(0.002)²

F = 4 x 10⁴ N

magnitude : 4 x 10⁴ N

Direction : North

we know that a negative charge pulls a positive charge towards it. the negative charge "q₁" is north of the positive charge. hence the charge q₁ apply force on charge q₂ in north direction.

so easy. just google the electromagnetic waves and it’ll pop up.

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min