Three charges are on the \(y\) -axis. Two of the charges, each \(-q,\) are located \(y=\pm d,\) and the third charge, \(+2 q,\) is located at \(y=0 .\) Derive an expression for the electric field at a point \(P\) on the \(x\) -axis.

The coordinates of the charges and point P are as follows: Charge 1: (-q) at (0,+d) Charge 2: (-q) at (0,-d) Charge 3: (+2q) at (0,0) Point P: (x,0)

We will use the electric field formula for a point charge: \(E=\dfrac{kq}{r^2}\), where k is the Coulomb's constant, q is the charge, and r is the distance from the charge to the point P. For Charge 1: P relative to Charge 1 will be at (x,-d) which makes \(r_1 = \sqrt{x^2 + d^2}\). Hence the electric field due to Charge 1 at Point P will be: \(E_1 = \dfrac{k(-q)}{(x^2 + d^2)}\) For Charge 2: P relative to Charge 2 will be at (x, d) which makes \(r_2 = \sqrt{x^2 + d^2}\). Hence the electric field due to Charge 2 at Point P will be: \(E_2 = \dfrac{k(-q)}{(x^2 + d^2)}\) For Charge 3: Since the Charge 3 is on the y-axis and P is on the x-axis, the distance between them is x. Hence, the electric field due to Charge 3 at Point P will be: \(E_3 = \dfrac{k(+2q)}{x^2}\)

Now, we need to find the components of the electric fields at point P in the x and y directions. For Charges 1 and 2, the electric fields will have both x and y components. Let's denote the electric field angles as θ1 and θ2. We find the angles through: \(cos(\theta_1) = \dfrac{x}{r_1}\) and \(cos(\theta_2) = \dfrac{x}{r_2}\) The x-components and y-components of electric fields at P due to Charges 1 and 2 will be: \(E_{1x} = E_1 cos(\theta_1) = \dfrac{k(-q)x}{(x^2 + d^2)^{\frac{3}{2}}}\) \(E_{1y} = E_1 sin(\theta_1) = \dfrac{k(-q)(-d)}{(x^2 + d^2)^{\frac{3}{2}}}\) \(E_{2x} = E_2 cos(\theta_2) = \dfrac{k(-q)x}{(x^2 + d^2)^{\frac{3}{2}}}\) \(E_{2y} = E_2 sin(\theta_2) = \dfrac{k(-q)d}{(x^2 + d^2)^{\frac{3}{2}}}\) Charge 3 produces an electric field only in the x-direction: \(E_{3x} = E_3 = \dfrac{k(+2q)}{x^2}\)

According to the principle of superposition, the net electric field at point P will be the vector sum of the electric fields due to all three charges. The net x-component of the electric field at P: \(E_x = E_{1x} + E_{2x} + E_{3x} = \dfrac{k(-q)x}{(x^2 + d^2)^{\frac{3}{2}}}+\dfrac{k(-q)x}{(x^2 + d^2)^{\frac{3}{2}}}+\dfrac{k(+2q)}{x^2} = \dfrac{-2kqx}{(x^2 + d^2)^{\frac{3}{2}}}+\dfrac{2kq}{x^2}\) Since \(E_{1y}\) and \(E_{2y}\) have opposite directions, they will cancel each other out: The net y-component of the electric field at P: \(E_y = E_{1y} + E_{2y} = 0\)

The net electric field at Point P on the x-axis is the vector sum of the x and y components: \(\vec{E} = E_x\hat{i} + E_y\hat{j} = \left(\dfrac{-2kqx}{(x^2 + d^2)^{\frac{3}{2}}}+\dfrac{2kq}{x^2}\right)\hat{i}\) Therefore, the electric field at point P on the x-axis due to the given charges is: \(\vec{E} = \left(\dfrac{-2kqx}{(x^2 + d^2)^{\frac{3}{2}}}+\dfrac{2kq}{x^2}\right)\hat{i}\)