Answer:
E = 781.12 N/C
Explanation:
Look at the attached graphic:
The 20µC charges are positive , then, the electric fields leave the charge.
The 22µC charges are negative, then, the electric fields enter the charge.
The electric field due to each of the charges is calculated by Coulomb's law:
E= k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
Problem development
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
E₁, E₂, E₃, E₄: Electric field at point P due to charge q₁, q₂, q₃, and q₄ respectively
The electric field in the direction of the y axis at the point P in the center of the square is equal to zero because the sum of the vertical components upwards is equal to the sum of the vertical components downwards.
Calculation of the electric field in the direction of the x-axis in the center of the square (point P)
Eₓ = E₁ₓ + E₂ₓ + E₃ₓ + E₄ₓ
E₁ₓ = E₂ₓ = (9*10⁹)*(20*10⁻⁶)*cos(45°)/(d²) = (9*10⁹)*(20*10⁻⁶)*cos(45°)/(26.16²) = 185.98 N/C
E₃ₓ = E₄ₓ = (9*10⁹)*(22*10⁻⁶)*cos(45°)/(d²) = (9*10⁹)*(22*10⁻⁶)*cos(45°)/(26.16²) = 204.58 N/C
Eₓ = E = 2*185.98 +2*204.58 = 781.12 N/C
Answer:
Explanation:
Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:
Where and
are the orbital periods of Mercury and Earth respectively. We have
and
. Replacing this and solving for
Answer:
1) Weight on Mercury
Explanation:
do the same to the rest and use your calculator to find the weight in N.