Answer:
Ep = 3797.05 N/C in the direction leaving the charge q₂ towards point P
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)
Data
k= 8.99*10⁹ N*m²/C²
q₁ = +13.6 x 10⁻⁹C
q₂ = +61.0*10⁻⁹C
d₁ =d₂= 33.5 cm = 0.335 m
Look at the attached graphic:
E₁: Electric Field at point P due to charge q₁. As the charge q₁ is positive (q₁+) ,the field leaves the charge
E₂: Electric Field at point P due to charge q₂. As the charge q₂ is positive (q₂+) ,the field leaves the charge
E₁ = k*q₁/d₁² = 8.99*10⁹ *13.6 *10⁻⁹/(0.335)² = 1089.45 N/C
E₂ = k*q₂/d₂²=- 8.99*10⁹ *61*10⁻⁹/(0.335)² = - 4886.5 N/C
Magnitude of the electric field at a point midway between q₁ and q₂
Ep= - 4886.5+ 1089.45 = -3797.05 N/C
Ep = 3797.5 N/C in the direction leaving the charge q₂ towards point P
Answer:
1. Primary or P waves are push and pull waves
2. Secondary, S or Shear Waves are also called transverse wave
3. L or surface waves reach the earth's surface after P and S waves
Answer:
a) ΔV = 2,118 10⁻⁸ m³ b) ΔR= 0.0143 cm
Explanation:
a) For this part we use the concept of density
ρ = m / V
As we are told that 1 carat is 0.2g we can make a rule of proportions (three) to find the weight of 2.8 carats
m = 2.8 Qt (0.2 g / 1 Qt) = 0.56 g = 0.56 10-3 kg
V = m / ρ
V = 0.56 / 3.52
V = 0.159 cm3
We use the relation of the bulk module
B = P / (Δv/V)
ΔV = V P / B
ΔV = 0.159 10⁻⁶ 58 10⁹ /4.43 10¹¹
ΔV = 2,118 10⁻⁸ m³
b) indicates that we approximate the diamond to a sphere
V = 4/3 π R³
For this part let's look for the initial radius
R₀ = ∛ ¾ V /π
R₀ = ∛ (¾ 0.159 /π)
R₀ = 0.3361 cm
Now we look for the final volume and with this the final radius
= V + ΔV
= 0.159 + 2.118 10⁻²
= 0.18018 cm3
= ∛ (¾ 0.18018 /π)
= 0.3504 cm
The radius increment is
ΔR = - R₀
ΔR = 0.3504 - 0.3361
ΔR= 0.0143 cm
Yea I would love to help u
