Topic of the Question:
This problem belongs to electrostatics, focusing on the properties of electric potential and electric field due to a point charge configuration, specifically an electric dipole system.
Step 1 : Understanding the Question:
We are given two equal and opposite point charges $+q$ and $-q$ separated by a fixed distance $d$. We need to evaluate both the net electric potential ($V$) and the net electric field ($E$) at the midpoint on the line joining these two charges.
Step 2 : Key Formulas and Approach:
Electric potential ($V$) is a scalar quantity. The net potential at a point is the algebraic sum of individual potentials: $V = \sum \frac{k q_i}{r_i}$, where $k = \frac{1}{4\pi\varepsilon_0}$.
Electric field ($\vec{E}$) is a vector quantity. The net electric field at any point is the vector sum of individual fields: $\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2 + \dots$, where the magnitude of a point charge's field is $E = \frac{k |q|}{r^2}$.
Step 3 : Detailed Explanation:
At the midway point, the distance from each point charge is equal to $r = \frac{d}{2}$.
To find the net electric potential, we sum the potential due to the positive charge and the potential due to the negative charge: $V_{\text{net}} = V_+ + V_- = \frac{k q}{d/2} + \frac{k (-q)}{d/2}$.
Simplifying this algebraic sum yields $V_{\text{net}} = \frac{2kq}{d} - \frac{2kq}{d} = 0$. Hence, the net electric potential at the center is exactly zero.
To calculate the net electric field, we analyze the direction of the field vectors. The electric field due to the positive charge ($\vec{E}_+$) points away from it, which is towards the negative charge (to the right).
The electric field due to the negative charge ($\vec{E}_-$) points towards it, which is also to the right. Since both vector fields point in the exact same direction, their magnitudes add constructively.
The total field magnitude is calculated as $E_{\text{net}} = E_+ + E_- = \frac{k q}{(d/2)^2} + \frac{k q}{(d/2)^2} = \frac{4kq}{d^2} + \frac{4kq}{d^2} = \frac{8kq}{d^2}$.
Since $q$ and $d$ are non-zero, this net electric field magnitude is clearly non-zero ($E \neq 0$).
Step 4 : Final Answer:
Based on the calculations, the net potential is zero ($V = 0$) and the net electric field is non-zero ($E \neq 0$). This matches the expressions provided in Option (D).