Question

A charge +3q is placed at x = 0 and a charge +11q is placed at x = 5 units. Where, along the x-axis is the net force on a charge Q equal to zero?

Answer 1

At x = 1.72 units  the net force on a charge Q equal to zero when a charge +3q is placed at x = 0 and a charge +11q is placed at x = 5 units.

A net force is defined as the sum of all the forces acting on an object. The equation below is the sum of N forces acting on an object. There may be several forces acting on an object, and when you add up all of those forces, the result is what we call the net force acting on the object.

This equation is the sum of n forces acting on an object. The magnitude of the net force acting on an object is equal to the mass of the object multiplied by the acceleration of the object, as shown in this formula.

+3q        ⇔        +11q

x = ?      ⇔         x = 5

$\frac{1}{4\pi E0} (\frac{3q}{x^{2} } -\frac{11q}{(5-x)^{2} } ) = 0$

$\frac{3}{x^{2} } = \frac{11q}{(5-x)^{2} }$

$\frac{5-x}{x} = \sqrt{\frac{11}{3} } = 1.91$

$5-x = 1.91x\\x = 1.715 = 1.72 units$

∴ At x = 1.72 units  the net force on a charge Q equal to zero

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