Answer:
The graph appears to be in error.
The actual figure appears to be a rhombus with sides of 5 and 15 with a height of 5
The work done (F * S) is the area of the rhombus
1/2 * (5 +15) * 5 = 50 J
Answer:
A. 
B. 
C. 
Explanation:
The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

is the capacitance,
is the common plate area,
is the plate separation and
is the permittivity of the material between the plates.
For air or free space,
is
called the permittivity of free space. In general,
where
is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum,
.
The energy stored in a capacitor is the average of the product of its charge and voltage.

Its charge,
, is related to its capacitance by
(this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for
,

A. Substituting for
in
,

B. When the distance is
,


C. When the distance is restored but with a dielectric material of dielectric constant,
, inserted, we have

Answer : Total energy dissipated is 10 J
Explanation :
It is given that,
Time. t = 10 s
Resistance of the resistors, R = 4-ohm
Current, I = 0.5 A
Power used is given by :

Where
E is the energy dissipated.
So, E = P t.............(1)
Since, 
So equation (1) becomes :



So, the correct option is (3)
Hence, this is the required solution.
Oxygen has<span> a higher electro negativity that then Sulfur, so Sulfur </span>will<span> " lose" electrons to Oxygen and that </span>is<span> the electrons </span>will be<span> pulled closer to the Oxygen causing, for oxygen to </span>have a negative<span> charge and the Sulfur to </span>have<span> a positive charge</span>