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 placed at:

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

 

Distance from each charge to point \( P(0, y) \) is:

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

Step 2: Electric Field Due to Each Charge

The magnitude of electric field due to a point charge is given by:

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

Let \( \vec{E}_+ \) and \( \vec{E}_- \) be the fields due to \( +q \) and \( -q \) respectively. They make an angle \( \theta \) with the vertical, where:

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

Because of symmetry:

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

 

Vertical component of the field due to one charge:

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

Net vertical electric field at point \( P \):

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

Step 3: At a Far-Off Point (\( y \gg a \))

If \( y \gg a \), then \( a^2 + y^2 \approx y^2 \). So,

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

Conclusion:

At large distances along the y-axis, the electric field due to the dipole varies as \( \frac{1}{y^2} \).