Over a period of more than 30 years, albert klein of california drove 2.5 × 106 km in one automobile. consider two charges, q1 =

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Over a period of more than 30 years, albert klein of california drove 2.5 × 106 km in one automobile. consider two charges, q1 =

2.0 c and q2 = 6.0 c, separated by klein's total driving distance. a third charge, q3 = 4.0 c, is placed on the line connect- ing q1 and q2. how far from q1 should q3 be placed for q3 to be in equilibrium

Physics

1 answer:

kicyunya [3.2K] · Answer 1 · 2026-02-28T01:21:30+03:00

For charge q3 to maintain equilibrium, the total force acting on it must equal zero.
Suppose we denote the total distance traveled by the car as L (this is for ease of understanding).
We can formulate a system of equations to derive a solution. Let’s assume the distance from q1 to q3 is r_1 and the distance from q3 to q2 is r_2, which must satisfy the equation:
$r_1+r_2=L km$
The second equation will represent the total force on charge q3:
$F_{q3}=F_{q3q1}+F_{q3q2}=0\\ 0=k_c\frac{q_1q_3}{r_1^2}+k_c\frac{q_3q_2}{r^2}$
k_c symbolizes Coulomb's constant. Given that the left side equals zero, we can divide the entire equation by k_c to eliminate it:
$0=\frac{q_1q_3}{r_1^2}+\frac{q_3q_2}{r^2}$
Next, we will solve this for r_1^2:
$0=\frac{8}{r_1^2}+\frac{24}{r^2}\\ \frac{1}{r_1^2}=-\frac{3}{r^2}\\ r_1^2=-\frac{r^2}{3};r_2=L-r_1\\ r_1^2=\frac{(L-r_1)^2}{3}\\ r_1^2=\frac{L^2-2Lr_1+r_1^2}{3}\\ 3r_1^2=L^2-2Lr_1+r_1^2\\ 2r_1^2+2Lr_1-L^2=0$
Now, we have a quadratic equation with the following parameters:
$a=2\\ b=2L\\ c=-L^2$
The two solutions can be expressed as:
$r_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{4L^2+8L^2}}{4}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{12L^2}}{4}\\$
We require a positive solution. When we input all values, we derive:
$r_1=0.915\cdot 10^6$km$