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Gravitation force =
= 26.3 * 
= 2.63 *
N
Gravitation force =
Using Hooke's law, Fspring=kΔx, find the elastic constant of a spring that stretches 2 cm. when a 4 Newton force is applied to i
1 answer:
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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
