Two charges Q₁ = +6 $\mu$C) and Q₂ = -2 $\mu$C) are brought from infinity to positions on the x-axis of x = -4 cm and x = +4 cm, respectively. Is it possible to bring a third charge Q₃ = +3 $\mu$C) from infinity to a point on the x-axis between the charges where the potential is zero, and if so, where would this position be?

Question

Quizplus · Accepted Answer

We can use the principle of superposition to find the position where the potential is zero. Let's start by finding the potential due to the first two charges at some point on the x-axis with coordinate x.

The potential due to the first charge is:

V1 = k*Q1*(x - (-4 cm))/r1      where r1 = sqrt((x - (-4 cm))^2)

V1 = k*6*10^-6*(x + 4)/sqrt((x + 4)^2)

Similarly, the potential due to the second charge is:

V2 = k*Q2*(x - 4 cm)/r2        where r2 = sqrt((x - 4 cm)^2)

V2 = k*(-2)*10^-6*(x - 4)/sqrt((x - 4)^2)

The total potential at the point is the sum of the potentials due to the individual charges:

Vtotal = V1 + V2

We want to find the value of x that makes Vtotal zero. We can solve for this by setting Vtotal equal to zero and solving for x:

0 = V1 + V2

0 = k*6*10^-6*(x + 4)/sqrt((x + 4)^2) + k*(-2)*10^-6*(x - 4)/sqrt((x - 4)^2)

We can simplify this by multiplying both sides by the denominators inside the square roots:

0 = k*6*10^-6*(x + 4)*(x - 4) + k*(-2)*10^-6*(x - 4)*(x + 4)

0 = k*10^-6*(x + 4)*(3*(x - 4) - (x + 4))

0 = (x + 4)*(3*(x - 4) - (x + 4))

0 = 2x - 8

x = 4 cm/2 = 2 cm

Therefore, the answer is choice C, x = +2 cm.

Two Charges Q1 = +6 μ\muμ C) and Q2 = -2 μ\muμ

Two Charges Q₁ = +6 $\mu$ C) and Q₂ = -2 $\mu$