Answer:
3.536*10^-6 C
Explanation:
The magnitude of the charge is expresses as Q = CV
C is the capacitance of the capacitor
V is the voltage across the capacitor
Get the capacitance
C = ε0A/d
ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m
A is the area = 0.2m²
d is the plate separation = 0.1mm = 0.0001m
Substitute
C = 8.84 x 10-12 * 0.2/0.0001
C = 1.768 x 10-8 F
Get the potential difference V
Using the formula for Electric field intensity
E = V/d
2.0 × 10^6 = V/0.0001
V = 2.0 × 10^6 * 0.0001
V = 2.0 × 10^2V
Get the charge on each plate.
Q = CV
Q = 1.768 x 10-8 * 2.0 × 10^2
Q = 3.536*10^-6 C
Hence the magnitude of the charge on each plate should be 3.536*10^-6 C
Answer:
Explanation:
Let c be the circumference and r be the radius
c = 2πr , r = c / 2π , area A = π r² = π (c/2π )² = (1/4π) x c²
flux (ψ) = BA = 1 X 1/4π X c²
dψ/dt = 1/4π x 2c dc/dt =1/2π x c x dc/dt
at t = 8 s
c = 161 - 13 x 8 = 57 cm , dc/dt = 13 cm/s
e = dψ/dt = (1 / 2π )x 57 x 13 x 10⁻⁴ = 118 x 10⁻⁴ V.
Answer:
A) B = 5.4 10⁻⁵ T, B) the positive side of the bar is to the West
Explanation:
A) For this exercise we must use the expression of Faraday's law for a moving body
fem = 
fem = 
- d (B l y) / dt = - B lv
B = 
we calculate
B = - 7.9 10⁻⁴ /(0.73 20)
B = 5.4 10⁻⁵ T
B) to determine which side of the bar is positive, we must use the right hand rule
the thumb points in the direction of the rod movement to the south, the magnetic field points in the horizontal direction and the rod is in the east-west direction.
Therefore the force points in the direction perpendicular to the velocity and the magnetic field is in the east direction; therefore the positive side of the bar is to the West
Answer:
0.5m
Explanation:
v=f×lamda
v is 300m/s, f is 600Hz, lamda is ?
lamda=v/f
lamda=300/600
lamda =3/6=1/2m
