❓What is the total electric field on the plane equidistant from two equal charges separated by a dist

What is the total electric field on the plane equidistant from two equal charges separated by a distance?

The two equal charges ($q$ at $x_3 = -d$ and $q$ at $x_3 = d$) create a total electric field ($\mathbf{E}$) on the equidistant plane ($x_3 = 0$) that is purely radial (in the $\mathbf{e}\rho$ direction): $\mathbf{E} = \mathbf{E}{q, \text{lower}} + \mathbf{E}{q, \text{upper}} = \frac{q \rho}{2\pi\epsilon_0 r^3} \mathbf{e}\rho$ , where $r^2 = \rho^2 + d^2$. Since the electric field has no component in the $x_3$ direction ($\mathbf{E}_3 = 0$), the field lines are parallel to the surface at this mid-plane.

The key to understanding the total electric field $\mathbf{E}$ on the equidistant plane ($x_3=0$) is the principle of superposition and symmetry.

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🧲 Total Electric Field on the Equidistant Plane

When two identical point charges ($q$) are placed symmetrically on the $x_3$-axis (at $x_3 = -d$ and $x_3 = d$), the total electric field $\mathbf{E}$ at any point on the plane $x_3 = 0$ is the vector sum of the fields from the individual charges: $\mathbf{E} = \mathbf{E}{\text{lower}} + \mathbf{E}{\text{upper}}$.

1. The Individual Fields

Consider a point $P$ on the $x_3=0$ plane, located a radial distance $\rho$ from the $x_3$-axis. The distance from each charge to $P$ is $r = \sqrt{\rho^2 + d^2}$.

• The field $\mathbf{E}_{\text{lower}}$ from the charge at $-d$ points away from the charge, towards the upper right.

• The field $\mathbf{E}_{\text{upper}}$ from the charge at $+d$ points away from the charge, towards the lower right.

Each field vector can be broken into two components: a radial component ($\mathbf{E}_{\rho}$) and an axial component.

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