Answer:
The maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
Explanation:
Given that,
dielectric constant k = 5.5
the area of each plate, A = 0.034 m²
separating distance, d = 2.0 mm = 2 x 10⁻³ m
magnitude of the electric field = 200 kN/C
Capacitance of the capacitor is calculated as follows;
Maximum potential difference:
V = E x d
V = 200000 x 2 x 10⁻³ = 400 V
Maximum energy that can be stored in the capacitor:
E = ¹/₂CV²
E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²
E = 6.62 x 10⁻⁵ J
Therefore, the maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
Total work done is 0.13 Joules
<h3>What is work done ?</h3>
The sum of the displacement and the component of the applied force of the object in the displacement direction is the work done by a force.
According to the given information
We need to find the work done
work done = force × distance
We are given,
force = 26 N
Distance = 0.0005 meter
hence ,
Work done = 26 × 0.005
= 0.13 Joules
Total work done is 0.13 Joules
To know more about Work done
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Which of the following statements is TRUE?
A. Pollen, cat hair, and mold could trigger an asthma attack for some people.
I am fairly certain that this one is correct. Hope this helps! :)
Answer:
36g
Explanation:
Given parameters:
Number of moles of H₂O = 2moles
Unknown:
Mass of H₂O = ?
Solution:
To solve this problem, use the expression below:
Mass of H₂O = number of moles x molar mass
Molar mass of H₂O = 2(1) + 16 = 18g/mol
Mass of H₂O = 2 x 18 = 36g
Answer:
Explanation:
The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of -40 nC is removed from the water droplet.
The charge on one electron, 
Let the N number of electrons have charge -40 nC, such that,
Now, mass of one electron = 
Therefore, mass of N electrons = 
It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.
Let it is m times the total mass of the droplet which is 
Then,
It is the required fraction of mass of the droplet.
