A spherical conductor is inside a uniform electric field with a magnitude of 1000 N/C to the right. What is the magnitude and di
rection of the electric field produced by the charges of the conductor? 2. What is the net electric field value inside a conductor? Explain using a diagram 3. You place both leads of a volt meter on the conductor from problem 1. What voltage does the meter read? Why?
1 answer:
Answer:
1.-E=1000N/C to the LEFT
2.-The electric field inside a conductor in electrostatic state is always zero (conductor proprieties).
3.-The voltmeter read 0V as differential voltage between two points from the conductor
Explanation:
1.The electric field inside the conductor must be zero (conductor proprieties). Then the charges create a electric field equal an opposite to the external electric field. In other words E=1000N/C to the LEFT
2. The electric field inside a conductor in electrostatic state is always zero. As shown in the figure the electric field induced by the charges in the sphere surface cancelled the EXTERN electric field.
3.If the Electric field inside the conductor is zero, that means that the Voltage in the hole conductor is constant (conductor proprieties). In other words the the voltmeter read 0v as differential voltage between two points from the conductor.
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Answer:
a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J
Explanation:
Let's find the capacitance of the capacitor
C =
C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³
C = 2.62 10⁻¹² F
for the initial data let's look for the accumulated charge on the plates
C =
Q₀ = C ΔV
Q₀ = 2.62 10⁻¹² 8.70
Q₀ = 22.8 10⁻¹² C
a) we look for the capacity for the new distance
C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³
C₁ = 1.04 10⁻¹² F
C₁ = Q₀ / ΔV₁
ΔV₁ = Q₀ / C₁
ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²
ΔV₁ = 21.9 V
b) initial stored energy
U₀ =
U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)
U₀ = 99.2 10⁻¹² J
c) final stored energy
U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)
U_f = 249.9 10⁻¹² J
d) the work of separating the plates
as energy is conserved work must be equal to energy change
W = U_f - U₀
W = (249.2 - 99.2) 10⁻¹²
W = 150 10⁻¹² J
note that as the energy increases the work must be supplied to the system