Answer:
A. We have that radius r = 4.00m intensity I = 8.00 W/m^
total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W
b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2
c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J
Answer:
Found in the nucleus, Has mass of one amu
Answer: 39.8 μC
Explanation:
The magnitude of the electric field generated by a capacitor is given by:

d is the distance between the plates.
For a capacitor, charge Q = CV where C is the capacitance and V is the voltage.

where A is the area of the plate and ε₀ is the absolute permittivity.
substituting, we get

It is given that the magnitude of the electric field that can exist in the capacitor before air breaks down is, E = 3 × 10⁶ N/C.
radius of the plates of the capacitor, r = 69 cm = 0.69 m
Area of the plates, A = πr² = 1.5 m²
Thus, the maximum charge that can be placed on disks without a spark is:
Q = E×ε₀×A
⇒ Q = 3 × 10⁶ N/C × 8.85 × 10⁻¹² F/m × 1.5 m² = 39.8 × 10⁻⁶ C = 39.8 μC.