Answer:

Explanation:
As we know that

E = 1.8 kV/mm
now we know that electric field between the plated of capacitor is given as

now we will have



now we have

now we have area of the plates of capacitor



A) When a charge is moved in an electric field the work done (W) is calculated as charge*(change in potential). We can write W = q*V or V = W/q = 10/1 = 10V . This voltage is a difference in electric potential between 2 points within the field. If the charge is positive, and positive work is done upon it, then the final position is more positive than the original one.
<span>b) If a charge (Q) is released from rest and falls through a potential difference V, then its gain in energy (KE if no other force acts on the charged body) is q*V = 10J. This is the same as the work done in moving the charge to its new position in part (a), and is an example of the conservation of energy.</span>
Oxygen is a product of photosynthesis, therefor lover levels of C02 would be in the area with more water due to more phtosynthetic processes happening converting it into oxygen.
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Answer:
0.8
Explanation:
The two spheres have the same potential, V.
Let the radius of the larger sphere be R and the radius of the smaller sphere be r,
=> R = 4r
Let the charge on the smaller sphere be q. Hence, the larger sphere will have charge Q - q.
The potential of the smaller sphere will be:

The potential of the larger sphere will be:

Inputting R = 4r,

Since
,

=> Q - q = 4q
=> 5q = Q
q = 0.2Q
The fraction of the charge Q that rests on the smaller sphere is 0.2
The charge of the larger sphere is:
Q - q = Q - 0.2Q = 0.8Q
∴ The fraction of the total charge Q that rests on the larger sphere is 0.8