1) 
The capacitance of a parallel-plate capacitor is given by:

where
is the vacuum permittivity
A is the area of each plate
d is the distance between the plates
Here, the radius of each plate is

so the area is

While the separation between the plates is

So the capacitance is

And now we can find the energy stored,which is given by:

2) 0.71 J/m^3
The magnitude of the electric field is given by

and the energy density of the electric field is given by

and using
, we find

2.7549 x 10^4 is the answer I hope this helped u
Answer:
1.92 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 200 Kg
Spring constant (K) = 10⁶ N/m
Workdone =?
Next, we shall determine the force exerted on the spring. This can be obtained as follow:
Mass (m) = 200 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = m × g
F = 200 × 9.8
F = 1960 N
Next we shall determine the extent to which the spring stretches. This can be obtained as follow:
Spring constant (K) = 10⁶ N/m
Force (F) = 1960 N
Extention (e) =?
F = Ke
1960 = 10⁶ × e
Divide both side by 10⁶
e = 1960 / 10⁶
e = 0.00196 m
Finally, we shall determine energy (Workdone) on the spring as follow:
Spring constant (K) = 10⁶ N/m
Extention (e) = 0.00196 m
Energy (E) =?
E = ½Ke²
E = ½ × 10⁶ × (0.00196)²
E = 1.92 J
Therefore, the Workdone on the spring is 1.92 J
Answer:
For cast iron we have

For copper

For Lead

For Zinc

Explanation:
As we know that final speed of the block is calculated by work energy theorem

now we have

now we have


For cast iron we have


For copper


For Lead


For Zinc

