Answer:
density = 5520 kg/m^3
Explanation:
given that
radius of earth = 6378 km
G = 6.67 x 10⁻¹¹ m³/kg.s²
g = 9.80 m/s²
we know,
mass of earth
M = 5.972 x 10²⁴ kg
density =
V = volume of the earth = 4/3πr³
V = 4/3 x 3.14 x (6378 x 10³)³
V = 1.08 x 10²¹ m³
density =
density = 5.52 x 10³ kg/m^3
density = 5520 kg/m^3
Answer:
Ep= 3.8 10⁵ N/C
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
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)
Equivalence
1nC= 10⁻⁹C
1cm= 10⁻²m
Data
k= 9*10⁹ N*m²/C²
q₁ =+7.5 nC = +7.5*10⁻⁹C
q₂ = -2.0 nC = -2.0*10⁻⁹C
d₁ =d₂ = 1.5cm = 1.5 *10⁻²m = 0.015 m
Calculation of the electric fieldsat the midpoint (P) between the two charges
Look at the attached graphic:
E₁: Electric Field at point ;Due to charge q₁. As the charge q₁ is positive negative (q₁+), the field leaves the charge .
E₂: Electric Field at point : Due to charge q₂. As the charge q₂ is negative (q₂-) ,the field enters the charge
E₁ = k*q₁/d₁² = 9*10⁹ *7.5 *10⁻⁹/ ( 0.015 )² = 3*10⁵ N/C
E₂ = k*q₂/d₂²= 9*10⁹ *2*10⁻⁹/( 0.015 )² = 0.8*10⁵ N/C
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Ep= E₁ + E₂
Ep= 3*10⁵ N/C + 0.8*10⁵ N/C
Ep= 3.8 10⁵ N/C
Answer:
E = 1.1 10⁶ N / C
Explanation:
In this case they indicate that we can approximate the membrane as a parallel plate capacitor, we can use
E =
note that in this case the electric field created by each plate goes in the same direction, they are added
let's calculate
E =
E = 1.1 10⁶ N / C