2. Après avoir déterminé l'intervalle
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
To solve this problem it is necessary to apply the concepts related to heat exchange in the vegetable and water.
By definition the exchange of heat is given by
where,
m = mass
c = specific heat
= Change in temperature
Therefore the total heat exchange is given as
Our values are given as,
Total mass is = 200lb ,however the mass of solid vegetable and water is given as,
Replacing at our equation we have,
Therefore the heat removed is 22411.2 Btu