Question

(a) An electric field $\vec{E} = E_{0}\hat{i}$ exists in a region of space. Draw three equipotential surfaces in the region.

(b) Two point charges $-q$ and $+q$ are located at points $(-a, 0, 0)$ and $(a, 0, 0)$ respectively. Find the electrostatic potential at point $(x, 0, 0)$ where $x \gg a$.

Hint:

Equipotentials are planes perpendicular to E. For the axis of a dipole, add the two potentials and use x much greater than a.

Answer 1

Step 1: Understanding the Concept:
An equipotential surface is a surface on which every point has the same electric potential. The electric field is always perpendicular to such a surface. Part (b) needs the potential of two opposite charges at a large distance.

Step 2: Part (a), equipotentials for \( \vec{E} = E_0\hat{i} \):
The field is uniform and points along \( +x \). Equipotential surfaces must be perpendicular to it, so they are flat planes parallel to the \( yz \) plane.
Draw three equally spaced parallel planes at \( x = x_1,\ x_2,\ x_3 \) (with \( x_1 < x_2 < x_3 \)), perpendicular to the field lines.
The potential falls along the field, since \( dV = -E_0\,dx \). So \( V_1 > V_2 > V_3 \), and equal spacing gives equal potential steps.

Step 3: Part (b), write the potential at the point:
The point \( (x,0,0) \) with \( x \gg a \) lies on the axis of the dipole. The charge \( +q \) is at distance \( x - a \), and \( -q \) is at distance \( x + a \).
\[ V = \dfrac{1}{4\pi\varepsilon_0}\left[\dfrac{q}{x-a} - \dfrac{q}{x+a}\right] \]

Step 4: Simplify:
\[ V = \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{(x+a)-(x-a)}{x^2-a^2} = \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{2a}{x^2-a^2} \]
For \( x \gg a \), we write \( x^2 - a^2 \approx x^2 \).
\[ V \approx \dfrac{2qa}{4\pi\varepsilon_0 x^2} = \dfrac{p}{4\pi\varepsilon_0 x^2} \]
Here \( p = 2qa \) is the dipole moment, directed from \( -q \) to \( +q \), that is, along \( +x \).

Final Answer:
Equipotentials are parallel planes perpendicular to \( x \). The potential on the axis at large \( x \) is \[ \boxed{V = \dfrac{2qa}{4\pi\varepsilon_0 x^2}} \]