Three point charges are arranged in a line. charge q 3 = +5 nc is at the origin. charge q 2 = -3 nc is located at x- +4.00 cm. charge q 1 is at x= + 2.00 cm. what is q 1, in magnitude and sign, if the net force on q3 is zero?

q1= +0.75 nC

Explanation:

As the electrostatic force is linear, we can apply the superposition principle to calculate the total force on q₃ due to q₂ and q1, according to Coulomb's Law, as follows:

F₃₂ = k*q₃*q₂/r₃₂² = 9*10⁹ N*m²/C²*+5nC*(-3 nC) / (0.04m)² = -84.4*10⁻⁶ N

F₃₁ = k*q₃*q₁ / r₃₁² = 9*10⁹ N*m²/C²*q₁*(+5 nC) / (0.02m)²

The total force on q₃ is just the sum of F₃₂ and F₃₁, which must add to 0, as follows:

F₃ = F₃₂ + F₃₁ = 0

⇒ -84.4* 10⁻⁶ N = -9*10⁹ N*m²/C²*q₁*(+5 nC) / (0.02m)²

Solving for q₁, we get:

q₁ = (84.4 / 11.25)*10⁻¹⁰ C = +0.75 nC

q₁ must be positive, in order to counteract the attractive force on q₃ due to q₂.

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