Answer:
Option (e)
Explanation:
A = 45 cm^2 = 0.0045 m^2, d = 0.080 mm = 0.080 x 10^-3 m,
Energy density = 100 J/m
Let Q be the charge on the plates.
Energy density = 1/2 x ε0 x E^2
100 = 0.5 x 8.854 x 10^-12 x E^2
E = 4.75 x 10^6 V/m
V = E x d
V = 4.75 x 10^6 x 0.080 x 10^-3 = 380.22 V
C = ε0 A / d
C = 8.854 x 10^-12 x 45 x 10^-4 / (0.080 x 10^-3) = 4.98 x 10^-10 F
Q = C x V = 4.98 x 10^-10 x 380.22 = 1.9 x 10^-7 C
Q = 190 nC
Answer:
No photo or graph is there
Explanation:
Answer:
12 nC
Explanation:
Capacity of the parallel plate capacitor
C = ε₀ A/d
ε₀ is constant having value of 8.85 x 10⁻¹² , A area of plate , d is distance between plate
Area of plate = π r²
= 3.14 x (0.8x 10⁻²)²
= 2 x 10⁻⁴
C = ( 8.85 X 10⁻¹² x 2 x 10⁻⁴ ) / 2.8 x 10⁻³
= 7.08 x 10⁻¹³
Potential difference between plate = field strength x distance between plate
= 6 x 10⁶ x 2.8 x 10⁻³
= 16.8 x 10³ V
Charge on plate = CV
=7.08 x 10⁻¹³ X 16.8 X 10³
11.9 X 10⁻⁹ C
12 nC .
Answer:
Explanation:
F = mω²R
F = 15(2π/8.5)²(7.8)
F = 63.93044788...
F = 63.9 N
answer a) is the closest. No idea how they got a value that low unless they used a poor approximation for π.