Question

Two point charges +q and −q are held at (a, 0) and (−a, 0) in x-y plane. Obtain an expression for the net electric field due to the charges at a point (0, y). Hence, find electric field at a far off point (y ≫ a).

Answer 1

Electric Field at Point \( P(0, y) \) due to a Dipole

Step 1: Distance from Point P to the Charges

Charges are located at:

• \( +q \) at \( (a, 0) \)
• \( -q \) at \( (-a, 0) \)

The distance from each charge to point \( P(0, y) \) is calculated as:

\[ r_{+} = r_{-} = \sqrt{a^2 + y^2} \]

Step 2: Electric Field from Each Charge

The magnitude of the electric field from a point charge is given by:

\[ E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{r^2} \]

Let \( \vec{E}_+ \) and \( \vec{E}_- \) represent the electric fields produced by \( +q \) and \( -q \), respectively. These fields form an angle \( \theta \) with the vertical axis, where:

\[ \theta = \tan^{-1}\left(\frac{a}{y}\right) \]

Due to symmetry:

• The horizontal components (\( E_x \)) of the electric fields cancel each other.
• The vertical components (\( E_y \)) of the electric fields add up.

 

The vertical component of the electric field from a single charge is:

\[ E_{y\pm} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q y}{(a^2 + y^2)^{3/2}} \]

The net vertical electric field at point \( P \) is the sum of these components:

\[ E_{\text{net}} = 2 E_{y\pm} = \frac{1}{2\pi\varepsilon_0} \cdot \frac{q y}{(a^2 + y^2)^{3/2}} \]

Step 3: Field at a Distant Point (\( y \gg a \))

When \( y \gg a \), the term \( a^2 + y^2 \) can be approximated as \( y^2 \). Therefore,

\[ E_{\text{far}} = \frac{1}{2\pi\varepsilon_0} \cdot \frac{q}{y^2} \]

Conclusion:

At significant distances along the y-axis, the electric field generated by the dipole diminishes proportionally to \( \frac{1}{y^2} \).