Two charges lie along the x-axis. The positive charge q1 = 15 µC is at x = 2 m, and the positive charge q2 = 6 µC is at the origin. Where must a negative charge (q3) be placed on t... Two charges lie along the x-axis. The positive charge q1 = 15 µC is at x = 2 m, and the positive charge q2 = 6 µC is at the origin. Where must a negative charge (q3) be placed on the x-axis so that the resultant electric force on it is zero? Read more

Steps to Solve

1. Identify the positions of the charges

The positive charges are located at ( x = 0 , m ) for ( q_2 = 6 , \mu C ) and ( x = 2 , m ) for ( q_1 = 15 , \mu C ).

2. Set the position of the negative charge

Let the position of the negative charge ( q_3 ) be ( x ) (the exact position along the x-axis is to be determined).

3. Determine the forces acting on ( q_3 )

The forces on ( q_3 ) will be due to both positive charges. The force exerted by ( q_2 ) will be attractive (pulling ( q_3 ) towards it) and will be given by Coulomb's Law:

$$ F_{2} = k \frac{ |q_2 q_3| }{ x^2 } $$

where ( k ) is Coulomb's constant.

The force exerted by ( q_1 ) will also be attractive and is given by:

$$ F_{1} = k \frac{ |q_1 q_3| }{ (x - 2)^2 } $$

4. Set the forces equal to each other for equilibrium

For the net force on ( q_3 ) to be zero, these two forces must be equal:

$$ F_{2} = F_{1} $$

Thus:

$$ k \frac{ |q_2 q_3| }{ x^2 } = k \frac{ |q_1 q_3| }{ (x - 2)^2 } $$

5. Cancel out constants and solve for ( x )

Since the constant ( k ) and ( |q_3| ) can be canceled out, we have:

$$ \frac{ |q_2| }{ x^2 } = \frac{ |q_1| }{ (x - 2)^2 } $$

Substituting the values of ( q_1 ) and ( q_2 ):

$$ \frac{ 6 \times 10^{-6} }{ x^2 } = \frac{ 15 \times 10^{-6} }{ (x - 2)^2 } $$

6. Cross-multiply and solve the equation

Cross-multiply to obtain:

$$ 6 \times 10^{-6} (x - 2)^2 = 15 \times 10^{-6} x^2 $$

Dividing through by ( 10^{-6} ):

$$ 6 (x - 2)^2 = 15 x^2 $$

Expanding and simplifying gives:

$$ 6 (x^2 - 4x + 4) = 15 x^2 $$

$$ 6x^2 - 24x + 24 = 15x^2 $$

$$ 0 = 9x^2 + 24x - 24 $$

7. Use the quadratic formula to find ( x )

The quadratic equation can be simplified to:

$$ 3x^2 + 8x - 8 = 0 $$

Using the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

where ( a = 3, b = 8, c = -8 ).

8. Calculate the discriminant and final solution

First, calculate the discriminant:

$$ b^2 - 4ac = 8^2 - 4 \cdot 3 \cdot (-8) = 64 + 96 = 160 $$

Now plug into the quadratic formula:

$$ x = \frac{-8 \pm \sqrt{160}}{2 \cdot 3} = \frac{-8 \pm 4\sqrt{10}}{6} = \frac{-4 \pm 2\sqrt{10}}{3} $$

Since we need a position on the x-axis, take the positive solution:

$$ x = \frac{-4 + 2\sqrt{10}}{3} $$