Problem 78 - A Complete Resource Book in Physics for JEE Main

Question

Two equal negative charges \(-q\) are fixed at points \((0, a)\) and \((0,-a)\) on the \(y\)-axis. A positive charge \(Q\) is released from rest at a point \((2 a, 0)\) on the \(x\)-axis. The charge \(Q\) will (A) execute SHM about the origin. (B) move to the origin and remain at rest there. (C) move to infinity. (D) execute oscillatory but not SHM.

Step-by-Step Solution

Answer

The motion of the positive charge \(Q\) will be (C) move to infinity due to the net electric force acting along the negative \(x\)-axis.

Step 1: Visualize the Problem

Draw the \(xy\)-plane with the charges positions. The two equal negative charges \(-q\) are placed symmetrically on the \(y\)-axis at points \((0, a)\) and \((0, -a)\). The positive charge \(Q\) is situated at the point \((2a, 0)\) on the \(x\)-axis.

Step 2: Determine Electric Force on \(Q\) from Charge \(-q\) at \((0, a)\)

To compute the electric force on charge \(Q\) due to charge \(-q\) at point \((0, a)\), we first calculate the distance between the charges: \(r_1 = \sqrt{(2a)^2 + a^2} = a\sqrt{5}\). Next, we apply Coulomb's law to find the electric force \(\vec{F}_{1}\) (magnitude and direction): \[\vec{F}_{1} = k\frac{|Q||q|}{r_1^2} \vec{u}_1\] Where \(k\) is Coulomb's constant and \(\vec{u}_1\) is the unit vector from charge \(Q\) to charge \(-q\) at \((0, a)\). We need to find the components of \(\vec{u}_1\). Let's assign \(\alpha\) as the angle between \(r_1\) and the negative \(x\)-axis. Using the tangent function: \[\tan \alpha = \frac{a}{2a} = \frac{1}{2}\] So, the components of \(\vec{u}_1\) are: \[\vec{u}_1 = -\cos\alpha\hat{i} - \sin\alpha\hat{j} = -\frac{2}{\sqrt{5}}\hat{i} - \frac{1}{\sqrt{5}}\hat{j}\] Substitute values to get the electric force \(\vec{F}_{1}\) acting on charge \(Q\) due to the charge at \((0, a)\).

Step 3: Determine Electric Force on \(Q\) from Charge \(-q\) at \((0, -a)\)

We perform a similar process as in Step 2 to find the electric force on charge \(Q\) due to the charge \(-q\) at point \((0, -a)\). Since the negative charges are symmetric, the distance between them and charge \(Q\) is the same \(r_2 = r_1 = a\sqrt{5}\). Let's define \(\beta\) as the angle between \(r_2\) and the negative \(x\)-axis. Using the tangent function: \[\tan \beta = \frac{-a}{2a} = -\frac{1}{2}\] So, the components of the unit vector \(\vec{u}_2\) are: \[\vec{u}_2 = -\cos\beta\hat{i} + \sin\beta\hat{j} = -\frac{2}{\sqrt{5}}\hat{i} + \frac{1}{\sqrt{5}}\hat{j}\] Substitute values to get the electric force \(\vec{F}_{2}\) acting on charge \(Q\) due to the charge at \((0, -a)\).

Step 4: Calculate Net Electric Force on \(Q\)

We now sum the electric forces \(\vec{F}_{1}\) and \(\vec{F}_{2}\) to find the net electric force \(\vec{F}_{net}\) on charge \(Q\). We can observe that the \(\hat{j}\) components of the forces have opposite signs and are equal in magnitude due to symmetry. This means that the \(\hat{j}\) components cancel each other out, and the net electric force is along the \(x\)-axis. Thus, \[\vec{F}_{net} = (\vec{F}_{1x} + \vec{F}_{2x})\hat{i} = -k\frac{2Qq}{5a^2}\hat{i}\]

Step 5: Analyze the Motion of \(Q\)

The net electric force on charge \(Q\) acts along the negative \(x\)-axis, implying that the charge moves to the left. The movement is not affected by the vertical positions of the negative charges since their forces cancel in the vertical direction due to symmetry. The motion of charge \(Q\) is not undergoing Simple Harmonic Motion (SHM), as there is no restoring force proportional to its displacement from an equilibrium position. Based on the analysis, the correct option is: (C) move to infinity.