Question

A charge \(Q_{1}\) is positioned on the \(x\) -axis at \(x=a\). Where should a charge \(Q_{2}=-4 Q_{1}\) be placed to produce a net electrostatic force of zero on a third charge, \(Q_{3}=Q_{1}\), located at the origin? a) at the origin c) at \(x=-2 a\) b) at \(x=2 a\) d) at \(x=-a\)

Short answer

Given that charge \(Q_1\) is located at \(x=a\) and \(Q_3\) is located at the origin. Answer: Charge \(Q_2\) should be positioned at both b) \(x = 2a\) and c) \(x = -2a\) on the x-axis to make the net electrostatic force on charge \(Q_3\) zero.

Step-by-step solution

Write down the given information and variables needed

: First, let's list the given information: - \(Q_1\) is located at \(x = a\) - \(Q_2\) has a charge of \(-4Q_1\) - \(Q_3\) is located at the origin, so its position is \((0,0)\) - \(Q_3\) has a charge of \(Q_1\) The position of \(Q_2\) is what we need to find, so we'll denote that as \(x = x_2\).

Determine the electrostatic force acting on \(Q_3\) due to \(Q_1\) and \(Q_2\)

: According to Coulomb's Law, the electrostatic force between two charges is given by: \(F = k\frac{Q_1Q_2}{r^2}\), where \(F\) is the electrostatic force, \(k\) is Coulomb's constant, and \(r\) is the distance between the charges. Therefore, the force on \(Q_3\) due to \(Q_1\) is: \(F_{13} = k\frac{Q_1Q_3}{(a - 0)^2} = k\frac{Q_1^2}{a^2}\) (since \(Q_3 = Q_1\)) And the force on \(Q_3\) due to \(Q_2\) will be: \(F_{23} = k\frac{Q_2Q_3}{(x_2 - 0)^2} = k\frac{-4Q_1^2}{x_2^2}\) (since \(Q_2 = -4Q_1\) and \(Q_3 = Q_1\)).

Apply the principle of superposition

: According to the principle of superposition, the net force on a charge is the vector sum of the forces due to other charges. In this case, we want the net force on \(Q_3\) to be zero, so \(F_{13} + F_{23} = 0\). Since both forces are acting along the x-axis, we can write the equation as: \(k\frac{Q_1^2}{a^2} + k\frac{-4Q_1^2}{x_2^2} = 0\)

Solve for \(x_2\)

: We can divide both sides of the equation by \(kQ_1^2\), which gives: \(\frac{1}{a^2} - \frac{4}{x_2^2} = 0\) Now, let's solve for \(x_2^2\): \(x_2^2 = \frac{4}{\frac{1}{a^2}}\) Now, multiply both sides by \(a^2\): \(x_2^2 = 4a^2\) Now, take the square root of both sides: \(x_2 = \pm 2a\) Since \(x_2\) has both positive and negative solutions, the answer is both b) at \(x = 2a\), and c) at \(x = -2a\).