Question

Two point charges $+q$ and $-q$ are held fixed at a distance $d$ apart. The net electric potential at a point midway between the two charges and its net field are?

• A. \( E=0 \quad V=0 \)
• B. \( E \neq 0 \quad V \neq 0 \)
• C. \( E=0 \quad V \neq 0 \)
• D. \( E \neq 0 \quad V=0 \)

Hint:

For a dipole system, at any point on the perpendicular bisector (including the midpoint), the net electric potential is always zero, while the net electric field is non-zero and parallel to the dipole axis.

Answer 1

The Correct Option is D

Concept: Electric potential (\(V\)) is a scalar quantity, so the total potential at a point is the simple algebraic sum of the potentials due to individual charges. \[ V_{\text{net}} = \sum V_i = \frac{k q_i}{r_i} \] Electric field (\(E\)) is a vector quantity, so the net electric field is the vector sum of individual fields, taking both magnitude and direction into account. \[ \vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2 + \dots \] Step 1: Calculating the net electric potential (\(V\)) at the midway point.
Let the distance of the midway point from both charges be \( r = \frac{d}{2} \). \[ V_{\text{net}} = V_+ + V_- = \frac{k q}{\frac{d}{2}} + \frac{k (-q)}{\frac{d}{2}} = \frac{2kq}{d} - \frac{2kq}{d} = 0 \]

Step 2: Calculating the net electric field (\(E\)) at the midway point.
The electric field due to the positive charge (\(+q\)) points away from it (towards the right). The electric field due to the negative charge (\(-q\)) points towards it (also towards the right). Since both field vectors point in the same direction, their magnitudes add up: \[ E_{\text{net}} = E_+ + E_- = \frac{k q}{\left(\frac{d}{2}\right)^2} + \frac{k q}{\left(\frac{d}{2}\right)^2} = \frac{4kq}{d^2} + \frac{4kq}{d^2} = \frac{8kq}{d^2} \neq 0 \]