Question

What is the net force and its direction that the charges at the vertices A and C of the right triangle ABC exert on the charge in vertex B?

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Answer 1

$\huge\underline\mathfrak{\pink{Solution}}$

Let FAB be the force of repulsion exerted by the charge at A on the charge at B and FCB be the force exerted by the charge at point C on the charge at point B.

The diagram below shows the direction of these two forces. We first use Coulomb's law (F = k q1 q2 / r2) to find the magnitude of these two forces

$\red{|FAB| = k (7 × 10-6)((2× 10-6) / (4 × 10-2)2 = 14 × 10-12 k / (16 × 10-4) = 0.875 k × 10-8</p><p></p><p>|FCB| = k (2× 10-6)(2× 10-6) / (2 × 10-2)2 = k × 10-8}$

We now use Pythagora's theorem to find the magnitude of the resultant force F = FAB + FCB (a vector sum)

$\red{|F| = √(|FAB|2 + |FCB|2) = k × 10-8 √( 0.875 2 + 12 ) = 9.00 × 109 × 10-8 √( 0.875 2 + 12 ) = 1.20 × 102 N</p><p></p><p>θ = arctan(|FCB|/ |FAB| ) = arctan( k × 10-8 / 0.875 k × 10-8) = 48.8°}$

Answer 2

Explanation:

Let FAB be the force of repulsion exerted by the charge at A on the charge at B and FCB be the force exerted by the charge at point C on the charge at point B.

The diagram below shows the direction of these two forces. We first use Coulomb's law (F = k q1 q2 / r2) to find the magnitude of these two forces

\red{|FAB| = k (7 × 10-6)((2× 10-6) / (4 × 10-2)2 = 14 × 10-12 k / (16 × 10-4) = 0.875 k × 10-8 |FCB| = k (2× 10-6)(2× 10-6) / (2 × 10-2)2 = k × 10-8}∣FAB∣=k(7×10−6)((2×10−6)/(4×10−2)2=14×10−12k/(16×10−4)=0.875k×10−8∣FCB∣=k(2×10−6)(2×10−6)/(2×10−2)2=k×10−8

We now use Pythagora's theorem to find the magnitude of the resultant force F = FAB + FCB (a vector sum)

\red{|F| = √(|FAB|2 + |FCB|2) = k × 10-8 √( 0.875 2 + 12 ) = 9.00 × 109 × 10-8 √( 0.875 2 + 12 ) = 1.20 × 102 N θ = arctan(|FCB|/ |FAB| ) = arctan( k × 10-8 / 0.875 k × 10-8) = 48.8°}∣F∣=√(∣FAB∣2+∣FCB∣2)=k×10−8√(0.8752+12)=9.00×109×10−8√(0.8752+12)=1.20×102Nθ=arctan(∣FCB∣/∣FAB∣)=arctan(k×10−8/0.875k×10−8)=48.8°