BILAL S. answered 05/23/23
Helping Students Master AP Physics & Math with Confidence
Hello Emalia;
Q1 and Q3 are at a distance ℓ to Q2 whereas Q4 is ℓ√2 far.
All are positive, all push Q2 away.
F(12) {Force of Q1 on Q2} = k8Q2/ℓ2 to the right (î) Let's temporarily name it 8F
F(32) {Force of Q3 on Q2} = k 72Q2/ℓ2 to upward (j) Let's temporarily name it 72F
F(42) {Force of Q3 on Q2} = k48Q2/(ℓ√2)2 = k24Q2/ℓ2. Let's temporarily name it 24F = (F(42)x12√2 right and F(42)y12√2 up) To add up components, I resolved this diagonal force into its x and y components.
Total X-dimension net force = 8F + 12√2F = F(8 + 12√2) (î)
Total Y-dimension net force = 72F + 12√2F = F(72 + 12√2) (j)
Using Pythagorean theorem;
F(net) =∼ 92.4F = 92.4kQ2/ℓ2
b) We need to find arc tan((72 + 12√2) / (8 + 12√2)) =≅74.3°
I hope this helps Emalia, numbers were not super friendly....