Three charged particles are situated on the x-axis. A charge of q is fixed at the origin, a charge of 3q is fixed at x = 12 cm, and a third charge, +q, is placed between q and 3q.... Three charged particles are situated on the x-axis. A charge of q is fixed at the origin, a charge of 3q is fixed at x = 12 cm, and a third charge, +q, is placed between q and 3q. Determine the position at which +q experiences a net force of zero. Read more

Steps to Solve

1. Understanding Charge Positions

   The charges are positioned as follows:

   • Charge $q$ at the origin (position $x = 0$ cm)
   • Charge $3q$ at position $x = 12$ cm
   • Charge $+q$ is placed at position $x = d$ cm, where $0 < d < 12$

2. Apply Coulomb's Law

   The force between two point charges is given by Coulomb's law: $$ F = k \frac{|q_1 q_2|}{r^2} $$ Where:

   • $F$ is the force between the charges
   • $k$ is Coulomb's constant
   • $q_1$ and $q_2$ are the magnitudes of the charges
   • $r$ is the distance between the charges

3. Calculate Forces Acting on +q

   The forces acting on charge $+q$ due to the other two charges are:

   • Force due to charge $q$: $$ F_{q \to +q} = k \frac{q \cdot q}{(d - 0)^2} = k \frac{q^2}{d^2} $$
   • Force due to charge $3q$: $$ F_{3q \to +q} = k \frac{3q \cdot q}{(12 - d)^2} = k \frac{3q^2}{(12 - d)^2} $$

4. Set Forces Equal for Net Force of Zero

   For the net force on charge $+q$ to be zero, the magnitudes of the forces must be equal: $$ F_{q \to +q} = F_{3q \to +q} $$ This gives us the equation: $$ k \frac{q^2}{d^2} = k \frac{3q^2}{(12 - d)^2} $$

5. Solve for d

   Dividing both sides by $kq^2$ (which is non-zero), we have: $$ \frac{1}{d^2} = \frac{3}{(12 - d)^2} $$ Cross-multiplying gives: $$ (12 - d)^2 = 3d^2 $$ Expanding and rearranging gives: $$ 144 - 24d + d^2 = 3d^2 $$ $$ 2d^2 + 24d - 144 = 0 $$ Dividing through by 2: $$ d^2 + 12d - 72 = 0 $$

6. Apply the Quadratic Formula

   To find the values of $d$, we can use the quadratic formula: $$ d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Where $a = 1$, $b = 12$, and $c = -72$: $$ d = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot (-72)}}{2 \cdot 1} $$ Simplifying: $$ d = \frac{-12 \pm \sqrt{144 + 288}}{2} $$ $$ d = \frac{-12 \pm \sqrt{432}}{2} $$ $$ \sqrt{432} = 12\sqrt{3} $$ Thus: $$ d = \frac{-12 \pm 12\sqrt{3}}{2} $$ Simplifying further: $$ d = -6 \pm 6\sqrt{3} $$

   We discard the negative solution since $d$ must be positive: $$ d = -6 + 6\sqrt{3} $$

7. Calculate the Numerical Value of d

   Estimating $d$ using $\sqrt{3} \approx 1.732$: $$ d \approx -6 + 10.392 \approx 4.392 \text{ cm} $$