Apositive charge, q1, of 5 µc is 3 × 10–2 m west of a positive charge, q2, of 2 µc. what is the magnitude and direction of the electrical force, fe, applied by q1 on q2? a) magnitude: 3 n direction: east b) magnitude: 3 n direction: west c) magnitude: 100 n direction: east d) magnitude: 100 n direction: west

The magnitude and direction of the electrical force, Fe, applied by q1 on q2 is:

C. magnitude: 100 N

direction: east

The correct answer option is C) magnitude: 100 N

, direction: east.

We know the formula of electrical force:

F = 1/4πε * q₁q₂ / r²

and we know the following values:

q₁ = 5 x 10⁻⁶ C

q₂ = 2 x 10⁻⁶ C

r = 3 x 10⁻² m west

So we can substitute these values in the formula to find the magnitude of F:

F = (9 x 10⁹) (5 x 10⁻⁶) (2 x 10⁻⁶)  / (3 x 10⁻²)²

F = 100 N  (East of positive charge)

Therefore, the magnitude of F is 100N applied by q1 and q2, which directs towards East of positive charge.

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multiply 2kg by latent heat of vap