Two charges ±10 μC are placed 5.0 mm apart.Determine the electric field at a point P on the axis of the dipole15 cm away from its centre O on the side of the positive charge, a point Q, 15 cm away from O on a linepassing through O and normal to the axis of the dipole

To determine the electric field at points P and Q due to a dipole consisting of charges ±10 μC separated by 5 mm, we will use the principles of electrostatics. A dipole consists of two equal and opposite charges separated by a distance, which creates an electric field around it. Let's break this down step by step.

Understanding the System

We have two point charges: +10 μC and -10 μC. They are separated by a distance of 5 mm. The electric field created by each charge can be calculated using Coulomb's Law, which states that the electric field (E) due to a point charge is given by:

E = k * |q| / r²

where:

• E = electric field
• k = Coulomb's constant (approximately 8.99 x 10⁹ N m²/C²)
• q = charge in coulombs
• r = distance from the charge

Position of the Charges and Points

Let’s define our points relative to the dipole:

• The center of the dipole (O) is the midpoint between the two charges.
• Point P is on the axis of the dipole, 15 cm from O on the side of the positive charge.
• Point Q is 15 cm away from O along a line that is perpendicular to the axis of the dipole.

Calculating the Electric Field at Point P

To find the electric field at point P, we need to calculate the contributions from both charges. The distance from O to P is 15 cm, and the distance from the positive charge to P is 10 cm (15 cm - 2.5 cm, since the distance between the charges is 2.5 cm from the center to each charge). The distance from the negative charge to P is 20 cm (15 cm + 2.5 cm).

Calculating the electric field due to the positive charge (+10 μC):

E₁ = k * |q₁| / r₁²

Here, r₁ = 0.1 m (10 cm),

E₁ = (8.99 x 10⁹) * (10 x 10⁻⁶) / (0.1)²

E₁ ≈ 8.99 x 10⁹ * 10 x 10⁻⁶ / 0.01 ≈ 8.99 x 10⁶ N/C

Calculating the electric field due to the negative charge (-10 μC):

E₂ = k * |q₂| / r₂²

Here, r₂ = 0.2 m (20 cm),

E₂ = (8.99 x 10⁹) * (10 x 10⁻⁶) / (0.2)²

E₂ ≈ 8.99 x 10⁹ * 10 x 10⁻⁶ / 0.04 ≈ 2.2475 x 10⁷ N/C

Since the electric field from the positive charge points away from the charge and towards the negative charge, the net electric field at P (E_P) will be the difference between E₂ and E₁:

E_P = E₂ - E₁ ≈ 2.2475 x 10⁷ - 8.99 x 10⁶ ≈ 1.3485 x 10⁷ N/C

Calculating the Electric Field at Point Q

For point Q, we will find the electric field contributions from both charges as well. The distance from O to Q is 15 cm, and the distances to each charge can be calculated using the Pythagorean theorem. The distance to the positive charge is:

r₁ = √((0.025)² + (0.15)²) = √(0.000625 + 0.0225) = √(0.023125) ≈ 0.1521 m

And the distance to the negative charge:

r₂ = √((0.025)² + (0.15)²) = 0.1521 m (same calculation)

Calculating the electric field due to the positive charge at Q:

E₁ = k * |q₁| / r₁² = (8.99 x 10⁹) * (10 x 10⁻⁶) / (0.1521)²

E₁ ≈ 3.88 x 10⁶ N/C

For the negative charge:

E₂ = k * |q₂| / r₂² = (8.99 x 10⁹) * (10 x 10⁻⁶) / (0.1521)²

E₂ ≈ 3.88 x 10⁶ N/C

Both electric fields will be in opposite directions, so the net electric field at Q (E_Q) is:

E_Q = E₁ + E₂ ≈ 3.88 x 10⁶ + 3.88 x 10⁶ = 7.76 x 10⁶ N/C

Summary of Results

In summary, we have calculated the electric fields at both points:

• Electric field at point P: approximately 1.3485 x 10⁷ N/C directed away from the dipole.
• Electric field at point Q: approximately 7.76 x 10⁶ N/C directed towards the dipole (since both fields point in the same direction).

This approach helps us understand how electric fields work in a dipole configuration and how distances affect the field strength. If you have further questions about electric fields or dipoles, feel free to ask!

Check the formula for electric field due to dipole at equatorial line & on axis.

                          Or

 You can find by vector components of Ë at req. points.         It's too easy!

               Good Luck!!!!!