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Question: The force between a pair of 0.005 C is 750 N. What is the distance between them?
Answer:
17.32 m
Explanation:
From coulomb's Law,
F = kqq'/r²........................... Equation 1
Where F = Force between the force, q' and q = both charges respectively, k = coulomb's constant, r = distance between both charges.
make r the subject of the equation above
r = √(kqq'/F)..................... Equation 2
From the question,
Given: q = q' = 0.005 C, F = 750 N
Constant: k = 9.0×10⁹ Nm²/C².
Substitute these values into equation 2
r = √(9.0×10⁹×0.005×0.005/750)
r = √(300)
r = 17.32 m.
Hence the distance between the pair of charges = 17.32 m
The new pressure P2 is 2.48 atmosphere.
<u>Explanation:</u>
Here, the one of the product of pressure and volume is equal to the products of pressure and volume of other.
By using Boyles's law,
pressure is inversely proportional to volume,
P1 V1 = P2 V2
where P1, V1 represents the first pressure and volume,
P2, V2 represents the second pressure and volume
P2 = (P1 V1) / V2
= (1.75 8.8) / 6.2
P2 = 2.48 atmosphere.