A circular surface with a radius of 0.058 m is exposed to a uniform external electric field of magnitude 1.49 104 N/C. The electric flux through the surface is 74 N · m2/C. What is the angle between the direction of the electric field and the normal to the surface?
To find the angle between the direction of the electric field and the normal to the surface, we can use the formula for electric flux:
Electric Flux = Electric Field * Area * Cos(θ)
Where:
- Electric Flux is given as 74 N·m²/C
- Electric Field is given as 1.49 × 10⁴ N/C
- Area of the circular surface = π * radius² (since it's a circle)
Let's calculate the area of the circular surface first:
Area = π * radius²
= π * (0.058 m)²
Next, we can rearrange the formula for electric flux to solve for the cosine of the angle:
Cos(θ) = Electric Flux / (Electric Field * Area)
Now, substitute the given values:
Cos(θ) = (74 N·m²/C) / (1.49 × 10⁴ N/C * π * (0.058 m)²)
Simplifying:
Cos(θ) = (74) / (1.49 × 10⁴ * π * 0.003364 m²)
Now, calculate the value of Cos(θ):
Cos(θ) ≈ 0.00109
To find the angle θ, we can take the inverse cosine (cos⁻¹) of this value:
θ ≈ cos⁻¹(0.00109)
Using a calculator, we find:
θ ≈ 89.97 degrees
Therefore, the angle between the direction of the electric field and the normal to the surface is approximately 89.97 degrees.