A circular surface with a radius of 0.053 m is exposed to a uniform external electric field of magnitude 1.48 104 N/C. The electric flux through the surface is 72 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 need to use the formula for electric flux:
Electric Flux = Electric Field * Area * cos(theta)
Given:
Radius of the circular surface (r) = 0.053 m
Magnitude of the electric field (E) = 1.48 * 10^4 N/C
Electric Flux (φ) = 72 N · m^2/C
We can rearrange the formula to solve for cos(theta):
cos(theta) = Electric Flux / (Electric Field * Area)
Finding the area of the circular surface:
Area = π * r^2
Area = 3.14159 * (0.053)^2
Area ≈ 0.00888 m^2
Now, substitute the values into the formula to find cos(theta):
cos(theta) = 72 / (1.48 * 10^4 * 0.00888)
cos(theta) ≈ 0.624
To find theta, take the inverse cosine (cos^-1) of 0.624:
theta ≈ cos^-1(0.624)
theta ≈ 51.06 degrees
Therefore, the angle between the direction of the electric field and the normal to the surface is approximately 51.06 degrees.
To find the angle between the direction of the electric field and the normal to the surface, we can use the definition of electric flux:
Electric Flux = Electric Field * Area * Cos(θ)
Where:
- Electric Flux is given as 72 N · m^2/C
- Electric Field is given as 1.48 x 10^4 N/C
- Area of the circular surface is given as πr^2, where r is the radius of the circular surface which is 0.053 meters
- θ is the angle between the direction of the electric field and the normal to the surface
Now, let's solve for θ.
First, let's substitute the given values into the formula:
72 N · m^2/C = (1.48 x 10^4 N/C) * (π * (0.053 m)^2) * Cos(θ)
Next, simplify and solve for Cos(θ):
Cos(θ) = (72 N · m^2/C) / [(1.48 x 10^4 N/C) * (π * (0.053 m)^2)]
Cos(θ) ≈ 0.335
Now, we can find the angle θ by taking the inverse cosine (arccos) of 0.335:
θ ≈ arccos(0.335)
Using a calculator, we find that θ ≈ 1.22 radians or ≈ 69.9 degrees.
Therefore, the angle between the direction of the electric field and the normal to the surface is approximately 69.9 degrees.