Question

Two point charges -q and +q are placed at a distance of L, as shown in the figure.

The magnitude of electric field intensity at a distance R (R>>L) varies as:

• A. \(\frac{1}{R^2}\)
• B. \(\frac{1}{R^3}\)
• C. \(\frac{1}{R^4}\)
• D. \(\frac{1}{R^6}\)

Answer 1

The Correct Option is B

Consider two point charges, -q and +q, placed at a distance \(L\) apart. We need to find the electric field intensity at a point at a large distance \( R \) such that \( R \gg L \). The electric field due to a point charge is given by:

\[ E = \frac{k \cdot |q|}{r^2} \]

where \( k \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge. For this system, we calculate the net electric field at a point at distance \( R \) from the midpoint between the charges. Since \( R \gg L \), the charges appear very close to each other relative to the distance \( R \). This system behaves like a dipole.

For a dipole, the electric field at a point along the perpendicular bisector at distance \( R \) is approximately given by:

\[ E_{\text{dipole}} = \frac{k \cdot p}{R^3} \]

where \( p = q \cdot L \) is the dipole moment. Thus, the electric field due to this configuration decreases with distance as:

\[ E \propto \frac{1}{R^3} \]

Therefore, the magnitude of the electric field intensity at a distance \( R \) varies as \(\frac{1}{R^3}\).