Question:

A) Two identical point charges each with a charge of magnitude +2.5 μC are separated by a distance of 0.5 m. Calculate the magnitude of the electric field at a point located 1.2 m away from the charges along the line connecting them.

B) Now, if one of the charges was replaced by a charge of magnitude -6 μC, what would be the magnitude and direction of the electric field at the same point mentioned in part A? Assume the positive direction is towards the charges.

Given:

• Charge of the point charges: +2.5 μC
• Distance between the charges: 0.5 m
• Distance from the charges to the point of interest: 1.2 m

Take the magnitude of the electric constant, ε₀, as 8.854 x 10⁻¹² C²/Nm².

Answer:

A) To calculate the magnitude of the electric field at the specified point, we can use the formula:

Electric field (E) = k * (q₁/r₁²) + k * (q₂/r₂²)

where:

• k is the electric constant (k ≈ 8.99 x 10⁹ Nm²/C²)
• q₁ and q₂ are the magnitudes of the point charges
• r₁ and r₂ are the distances between the charges and the point of interest

Substituting the known values into the formula:

r₁ = 0.5 m r₂ = 0.5 m q₁ = +2.5 μC = +2.5 x 10⁻⁶ C q₂ = +2.5 μC = +2.5 x 10⁻⁶ C

Electric field (E) = (8.99 x 10⁹ Nm²/C²) * [(2.5 x 10⁻⁶ C / (0.5 m)²) + (2.5 x 10⁻⁶ C / (0.5 m)²)]

Electric field (E) = (8.99 x 10⁹ Nm²/C²) * [(2.5 x 10⁻⁶ C / 0.25 m²) + (2.5 x 10⁻⁶ C / 0.25 m²)]

Electric field (E) = (8.99 x 10⁹ Nm²/C²) * [(2.5 x 10⁻⁶ C x 4) / (0.25 m²)]

Electric field (E) = (8.99 x 10⁹ Nm²/C²) * (10⁻⁵ C / 0.25 m²)

Electric field (E) = (8.99 x 10⁹ Nm²/C²) * (4 x 10⁻⁵ C / 0.25 m²)

Electric field (E) = (8.99 x 10⁹ Nm²/C²) * (4 x 10⁻⁵ C / 0.0625 m²)

Electric field (E) = (8.99 x 10⁹ Nm²/C²) * (0.000064 C / 0.0625 m²)

Electric field (E) = (8.99 x 10⁹ Nm²/C²) * (1.024 C/m²)

Electric field (E) = 9.2176 x 10⁶ N/C

Therefore, the magnitude of the electric field at the specified point along the line connecting the point charges is 9.2176 x 10⁶ N/C.

B) To calculate the new magnitude and direction of the electric field at the same point, we can use the same formula applied in part A, but now with the new charges:

r₁ = 0.5 m r₂ = 0.5 m q₁ = +2.5 μC = +2.5 x 10⁻⁶ C q₂ = -6 μC = -6 x 10⁻⁶ C

Electric field (E) = (8.99 x 10⁹ Nm²/C²) * [(2.5 x 10⁻⁶ C / (0.5 m)²) + (-6 x 10⁻⁶ C / (0.5 m)²)]

Electric field (E) = (8.99 x 10⁹ Nm²/C²) * [(2.5 x 10⁻⁶ C / 0.25 m²) + (-6 x 10⁻⁶ C / 0.25 m²)]

Electric field (E) = (8.99 x 10⁹ Nm²/C²) * [(2.5 x 10⁻⁶ C x 4) / (0.25 m²)] + [(6 x 10⁻⁶ C x 4) / (0.25 m²)]

Electric field (E) = (8.99 x 10⁹ Nm²/C²) * (10⁻⁵ C / 0.25 m²) + (2.4 x 10⁻⁵ C / 0.25 m²)

Electric field (E) = (8.99 x 10⁹ Nm²/C²) * (4 x 10⁻⁵ C / 0.25 m²) + (2.4 x 10⁻⁵ C / 0.25 m²)

Electric field (E) = (8.99 x 10⁹ Nm²/C²) * (4 x 10⁻⁵ C / 0.0625 m²) + (2.4 x 10⁻⁵ C / 0.0625 m²)

Electric field (E) = (8.99 x 10⁹ Nm²/C²) * (0.000064 C / 0.0625 m²) + (0.000384 C / 0.0625 m²)

Electric field (E) = (8.99 x 10⁹ Nm²/C²) * (1.024 C/m²) + (6.144 C/m²)

Electric field (E) = 9.2176 x 10⁶ N/C + 55.4912 N/C

Electric field (E) = 9.2737 x 10⁶ N/C

Therefore, the new magnitude of the electric field at the specified point along the line connecting the point charges when one of the charges is replaced with a charge of magnitude -6 μC is 9.2737 x 10⁶ N/C. The direction of the electric field remains the same since it is a scalar quantity.