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?
E normal=74*area
arccosTheta=Enormal/Eexternal
To find the angle between the direction of the electric field and the normal to the surface, you can use the formula:
Electric Flux = Electric Field * Area * cos(theta)
Where:
- Electric Flux is given as 74 N · m^2/C,
- Electric Field magnitude is given as 1.49 * 10^4 N/C, and
- Area of the circular surface is determined by its radius (r), which is given as 0.058 m (since it's a circular surface, the area is A = πr^2).
So, by rearranging the formula, we have:
cos(theta) = Electric Flux / (Electric Field * Area)
Let's substitute the given values into the formula to solve for the angle theta:
cos(theta) = 74 N · m^2/C / (1.49 * 10^4 N/C * π * (0.058 m)^2)
Now, we can calculate this value using a calculator:
cos(theta) ≈ 0.202
To find the angle theta, we can take the inverse cosine (cos^-1) of 0.202:
theta ≈ cos^-1(0.202)
Using a calculator or trigonometric table, the approximate angle theta is:
theta ≈ 78.1 degrees (rounded to one decimal place)
So, the angle between the direction of the electric field and the normal to the surface is approximately 78.1 degrees.