How this induced force can be increased? Write three ways.
1 answer:
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Answer:
1.6 pF
Explanation:
The capacitance of a parallel-plate capacitor in air is given by:
where
is the vacuum permittivity
A is the area of each plate
d is the separation between the plates
In this problem we have:
- Separation between the plates: d = 5.00 mm = 0.005 m
- Area of the plates:
Therefore, the capacitance is
I have attached the image showing the 6 combination of charges
Answer:
Smallest are: q1 = +1 nc, q2 = +1 nc, q3 = +1 nc and q1 = -1 nc, q2 = -1 nc, q3 = -1nc
Third Largest: q1 = +1 nc, q2 = +1nc, q3 = -1 nc
Second Largest: q1 = +1 nc, q2 = -1 nc, q3 = +1 nc
Largest are: q1 = +1 nc, q2 = -1 nc, q3 = -1 nc and q1 = -1 nc, q2 = +1 nc, q3 = +1 nc
Explanation:
The electrostatic force between 2 charges is; F = k•q1•q2/r²
Where r is the distance between the 2 charges q1 and q2 and k is coulombs constant.
Thus;
F ∝ q1•q2/r²
The net force on the charge q1 is the sum of the forces on charge q2 and q3.
If the two charges are opposite, the force is an attractive force while if the two charges are similar, the force is a repulsive force.
Thus, if the combination has same type of charges, i.e;(+1,+1,+1) or (-1,-1,-1), the force will be very small.
So, now the order of forces from smallest to largest is;
Smallest are: q1 = +1 nc, q2 = +1 nc, q3 = +1 nc and q1 = -1 nc, q2 = -1 nc, q3 = -1nc
Third Largest: q1 = +1 nc, q2 = +1nc, q3 = -1 nc
Second Largest: q1 = +1 nc, q2 = -1 nc, q3 = +1 nc
Largest are: q1 = +1 nc, q2 = -1 nc, q3 = -1 nc and q1 = -1 nc, q2 = +1 nc, q3 = +1 nc
