Question

Charge A and charge B are 3.00 m apart, and charge A is +1.44 C and charge B is +3.10 C. Charge C is located between them at a certain point and the force on charge C is zero. How far from charge A is charge C?

Answer 1

Answer:

the distance from charge A to C is r₁₃= 1.216 m

Explanation:

following Coulomb's law , the force exerted by 2 point charges between themselves is:

F= k*q₁*q₂/r₁₂² , where q is charge , r is distance and 1 and 2 represents the charge A and charge B respectively , k=constant

since C ( denoted as 3) is at equilibrium

F₁₃=F₂₃

k*q₁*q₃/r₁₃²=k*q₂*q₃/r₂₃²

q₁/r₁₃²=q₂/r₂₃²

r₁₃²/q₁=r₂₃²/q₂

r₂₃=r₁₃*√(q₂/q₁)

since C is at rest and is co linear with A and B ( otherwise it would receive a net force in either vertical or horizontal direction) , we have

r₁₃+r₂₃=d=r₁₂

r₁₃+r₁₃*√(q₂/q₁)=d

r₁₃*(1+√(q₂/q₁))=d

r₁₃=d/(1+√(q₂/q₁))

replacing values

r₁₃=d/(1+√(q₂/q₁)) = 3.00 m/(1+√(3.10 C/1.44 C)) = 1.216 m

thus the distance from charge A to C is r₁₃= 1.216 m

Answer 2

The distance between charge A and charge C is 1.22 m.

The given parameters:

• Distance between charge A and B = 3 m
• Charge A = 1.44 C
• Charge B = 3.1 C

If the net force on charge C is zero, then force due to charge A must be equal to the force due to charge B.

$F_A = F_B\\\\ \frac{kq_A q_C}{c^2 } = \frac{kq_B q_C}{(3-c)^2} \\\\ \frac{q_A}{c^2} = \frac{q_B}{(3-c)^2} \\\\ q_B c^2 = q_A(3-c)^2\\\\ 3.1c^2 = 1.44(3-c)^2\\\\ 3.1c^2 = 1.44(9 -6c + c^2)\\\\ 3.1c^2 = 12.96 -8.64c+ 1.44c^2$

$1.66c^2 + 8.64c - 12.96= 0\\\\ c = 1.22 \ m$

Thus, the distance between charge A and charge C is 1.22 m.

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