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
Answer with Explanation:
a. Option d is true.
a negatively charged plane parallel to the end faces of the cylinder
b. Radius of cylinder, r=0.66m
Magnitude of electric field, E=300 N/C
We have to find the net flux through the closed surface.
Net electric flux,
c.
Net charge,
Where
Where 
7) c
opposite charges attract, equal charges repell
8) c
Followed from equation
9) d
Same reason as under 7
Answer:
Explanation:
power of a heating element = V² / R where V is voltage and R is resistance
putting in the values
5200 = 75² / R
R = 1.082 ohm
Formula for resistance
R = ρ L / S , ρ is resistivity , L is length and S is cross sectional area
1.082 = 5 x 10⁻⁷ x L / (3.83 x 10⁻⁶)
L = 8.29 m
b )
power of a heating element = V² / R , R is new resistance
5200 = 120² / R
R = 2.77 ohm
R = ρ L / S , ρ is resistivity , L is length and S is cross sectional area
2.77 = 5 x 10⁻⁷ x L / (3.83 x 10⁻⁶)
L = 21.22 m
Answer:
The darts hit very different areas of the board.
Explanation:
