91Ó°ÊÓ

Let E1 be the field from the charge \(2q\) and E2 from \(-q\). Because the charges are on the x-axis, the direction of the electric fields will be along the x-axis.

The electric field from a point charge is given by \(E=k\frac{|q|}{r^2}\) where r is the distance from the charge. For \(E1\), \(q=2q\) and \(r=x\), resulting in \(E1=k\frac{2q}{x^2}\). For \(E2\), \(q=-q\) and \(r=x-a\), resulting in \(E2=-k\frac{q}{(x-a)^2}\). The negative sign comes from the negative charge.

Set \(E1 + E2 = 0\) and solve for \(x\). This gives: \(k\frac{2q}{x^2} = k\frac{q}{(x-a)^2}\). Simplifying this equation gives two roots, \(x=0, 2a\). However, \(x=0\) is the position of the \(2q\) charge and should be discarded. Thus, the electric field is zero at \(x=2a\) on the \(x\)-axis.

Here, the electric field is parallel to the x-axis when the y-component cancels out. At any point in the plane, the electric field has two components: one along the x-axis and one along the y-axis. The field from a charge always points directly away from that charge if positive or towards it if negative. Now, let’s consider a position \(P(a, y)\) on line \(x=a\). At this position, the x-component from \(2q\) cancels with the y-component from \(−q\), leaving only the y-component from \(2q\) and the x-component from \(-q\). For these components to cancel, they must be equal. So, \(k\frac{2q}{y^2} = k\frac{q}{y^2 + a^2}\). This results in \(y =±\frac{a}{\sqrt{3}}\), at least approximately. Note that the location along the \(y\)-direction should be positive because the electric field is parallel to the \(x\)-axis in the \(y>0\) region.