1. Analyzing the Potential $V$

Let $V_{q0}$ be the potential at P due to the bottom plate (charge $q_0$) and $V_q$ be the potential due to the top plate (charge $q$).

Since P is equidistant from both plates, the geometric factors relating charge to potential are identical for both plates. Thus, if $V_{q0} = k q_0$, then $V_q = k q$.

The total potential is:

$$ V(r) = V_{q0} + V_q = k q_0 + k q = k q_0 (1 + r) $$

where $r = q/q_0$.

From the graph, at $r = 1$ (where $q=q_0$), let the potential be $V_1$. Then $V_1 = 2k q_0$. Substituting $k q_0 = V_1/2$ back into the equation:

$$ V(r) = \frac{V_1}{2}(1 + r) $$

(Note: If the potential is defined as a potential difference across the plates, the dependence would be linear with charge difference, but the graph shows $V=0$ at $r=-1$, which implies superposition of potentials relative to a common zero, consistent with the derivation above.)

2. Analyzing the Electric Field $\vec{E}$

The electric field is a vector sum. Let $\vec{E}_{bot}$ be the field due to the bottom plate ($q_0$) and $\vec{E}_{top}$ be the field due to the top plate ($q$).

Due to the symmetry (P is in the mid-plane):

• Horizontal components ($x$): Both plates produce horizontal components in the same direction (bulging out). $\vec{E}_{x} \propto (q_0 + q)$.
• Vertical components ($y$): A positive bottom plate pushes up ($+y$). A positive top plate pushes down ($-y$). Thus, $\vec{E}_{y} \propto (q_0 – q)$.

Let the field magnitude at $r=1$ ($q=q_0$) be $E_1$. In this case, $q_0-q=0$, so the field is purely horizontal: $E_x = E_1$.

Let the field magnitude at $r=-1$ ($q=-q_0$) be $E_2$. In this case, $q_0+q=0$, so the field is purely vertical: $E_y = E_2$.

The general components for ratio $r$ are:

$$ E_x(r) = E_1 \frac{(q_0 + q)}{2q_0} = \frac{E_1}{2}(1+r) $$ $$ E_y(r) = E_2 \frac{(q_0 – q)}{2q_0} = \frac{E_2}{2}(1-r) $$

The modulus of the electric field $E = \sqrt{E_x^2 + E_y^2}$ is:

$$ E(r) = \sqrt{ \left[ \frac{E_1}{2}(1+r) \right]^2 + \left[ \frac{E_2}{2}(1-r) \right]^2 } $$