Two tiny particles having charges of +5.00E-6 C ...

Question

Two tiny particles having charges of +5.00E-6 C and +7.00E-6 C are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm, and the other particle is at x = 100.00 cm. Where on the x-axis must a third charged particle be placed so that it does not experience any net electrostatic force due to the other two particles?  Answer in cm

Answer 1

This is a 1-Dimensional force problem dealing with charged, stationary particles. First thing to do is list our known values.

q₁ = 5 × 10^-6 Coulombs = 5 μC (microCoulombs)

q₂ = 7 × 10^-6 Coulombs = 7μC

x₁ = 0cm

x₂ = 100cm = 1m

Because the charges are both positive, they will repel each other. Drawing our particles (+) with the forces (← or →) acting on each, our situation looks like this:

← + +→

Because these charges are positive, they will both either push or repel a 3rd charge. A good placement for a charge then would be somewhere in the middle of these two charges, so that the picture looks like this, with 0 net force on the central charge:

← + + +→

We want the forces on q3 to sum to 0. If q3 is positive, then q1 will push to the right in the positive direction and q2 will push left in the negative direction, giving the equation:

F_q3total = F_{q1 on q3} + F_{q2 on q3} = k*q₁q₃/r₁² - k*q₂q₃/r₂² = 0

(note, you could also assume that q3 was negative. The above equation would switch signs, but give the same result)

Here we used the equation for force (F =kq₁q₂/r²), where r₁ is the distance from q3 to q1 and r₂ the distance from q3 to q2, and k is the electric constant.

Once we know r1, the problem is finished. We know that r1 and r2 are related: since q1 and q2 are 1 meter apart, if q3 is r₁ distance away from q1, then it has to be 1meter minus r₁ distance away from q2. Rewriting our force equation gives (and letting r₁ = r)

F_q3total = k*q₁q₃/r² - k*q₂q₃/(1-r)² = kq₃(q₁/r² - q₂/(1-r)²) = 0

⇒ q₁/r² - q₂/(1-r)² = 0 (eliminate factor kq₃)

⇒((1-r)² q₁ - r²q₂)/[(1-r)² r²] = 0 (getting both terms under same denominator)

⇒[(r² -2r+1) q₁ - r²q₂)] = 0 (expanded (1-r)² and removed denominator: numerator is what goes to 0)

⇒(q₁ -q₂)r² -2q₁r+q₁ = 0 (collected like terms)

⇒ r = √q₁ / [√q₁ ± √q₂] (quadratic formula, only one answer is in the bounds we defined)

• r = √q₁ / [√q₁ + √q₂] (The distance from q3 to q1)

Because the question defined q1 to be at 0cm, the above will give you the final answer. Remember to use appropriate units (don't mix μC and C) and to report the final answer in cm (it should be slightly closer to q1 than q2)