Question

What is the magnitude of the electric field 29.7 cm directly above an isolated charge of 3.15×10^-5C? Provide a step-by-step explanation of the calculations involved in determining the electric field strength at this particular distance from the charge.

Answer 1

The magnitude of the electric field 29.7 cm above the charge is found using Coulomb's law and is approximately 3.210 × 10^6 N/C.

Step-by-step explanation:

Calculating Electric Field Magnitude

To calculate the magnitude of the electric field created by a point charge, we use Coulomb's law, specifically the electric field equation. The electric field (E) at a distance (r) from a point charge (q) is given by:

E = k * |q| / r^2

where:

• k is Coulomb's constant (8.99 × 10^9 N · m^2/C^2)
• q is the charge (3.15 × 10^-5 C)
• r is the distance from the charge (29.7 cm or 0.297 m)

The magnitude of the electric field strength can then be calculated by inserting these values into the equation:

E = (8.99 × 10^9 N · m^2/C^2) × (3.15 × 10^-5 C) / (0.297 m)^2

Performing the calculation:

E = (8.99 × 10^9) × (3.15 × 10^-5) / 0.297^2

E = (2.8321 × 10^5) / 0.088209

E = 3.210 × 10^6 N/C

Thus, the magnitude of the electric field 29.7 cm directly above the charge is approximately 3.210 × 10^6 N/C.