1- A 40.0cm diameter loop is turned to the position where the largest flux is handled in a uniform electric field. In this position, the flux is obtained as 5.20 × 10 ^ 5 N.m ^ 2/c Find the twist of the electric field.
2- An electron was released from rest in a uniform electric field of 5.90 × 10 ^ 3 V / m. How much potential awareness (V) will pass after 1cm movement, and what is the time of the electron in position (m / s)?
[m (e) = 9.11 × 10 ^ -31 kg, | q (e) | = 1.60 × 10 ^ -9 C.]
To answer these questions, we need to apply the principles of electric flux and electric potential. Let's break down each question step by step:
1. To find the twist of the electric field, we need to calculate the electric flux. The electric flux φ is given by the equation: φ = E * A * cos(θ), where E is the electric field strength, A is the area of the loop, and θ is the angle between the electric field and the normal to the loop.
In this case, the electric flux is given as 5.20 × 10^5 N.m^2/C, and the diameter of the loop is 40.0 cm. Recall that the diameter is twice the radius, so the radius of the loop is 20.0 cm or 0.20 m.
The area of the loop can be calculated using the formula: A = π * r^2, where r is the radius. Therefore, A = π * (0.20 m)^2.
Next, we need to find the electric field strength E. Rearranging the flux equation, E = φ / (A * cos(θ)).
Since the loop is turned to the position where the largest flux occurs, the angle θ will be 0°. Therefore, cos(θ) = cos(0°) = 1.
Substituting the given values, we can calculate E = (5.20 × 10^5 N.m^2/C) / (π * (0.20 m)^2).
2. To find the potential awareness (V) and the time of the electron in position (m/s), we need to use the equation for electric potential energy. The electric potential energy ΔPE of a charged object moving in an electric field is given by ΔPE = q * ΔV, where q is the charge and ΔV is the potential difference.
In this case, we have an electron with charge q(e) = 1.60 × 10^(-9) C and a uniform electric field with an electric potential of 5.90 × 10^3 V/m. The potential difference ΔV is given by ΔV = E * d, where E is the electric field strength and d is the distance traveled by the electron.
Since the electron moves 1 cm, we convert it to meters: d = 0.01 m.
Substituting these values, we can calculate ΔV = (5.90 × 10^3 V/m) * (0.01 m).
To find the time it takes for the electron to move, we can use the equation for velocity: v = d / t, where v is the velocity, d is the distance, and t is the time.
Rearranging the equation, we have t = d / v.
The velocity of the electron can be calculated using the equation v = a * t, where a is the acceleration and t is the time. Since the electron starts from rest, the initial velocity is 0.
The acceleration of the electron can be calculated using the equation a = F / m, where F is the force on the electron and m is the mass of the electron.
The force on the electron can be calculated using the equation F = q * E, where q is the charge of the electron and E is the electric field strength.
Substituting the given values, we can calculate F = (1.60 × 10^(-9) C) * (5.90 × 10^3 V/m).
Finally, we can calculate the acceleration a = F / m = (1.60 × 10^(-9) C) * (5.90 × 10^3 V/m) / (9.11 × 10^(-31) kg).
Now, with the acceleration calculated, we can find the time t = d / v = 0.01 m / (a * t).
By solving these equations, we can find the potential awareness (V) and the time of the electron in position (m/s).
Remember to substitute the values correctly and include all available units to ensure accurate calculations!