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Explain Air column and Fluid column.?
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according to me. , the coulomb force between two charges q1, q2 is:
F =kq1q2/r^2
k = coulomb’s constant = 1/4πεε0
ε0 = permittivity of empty space
ε = relative permittivity of the material in which the charges are immersed
r = distance between charges
for vacuum ε=1 and for air also, to a very good approximation ε=1.
For water , at room temperature ( 20C) ε = 80.
Therefore, the force between the charges q1, q2 will be 1/80 its value in air.
Hence Fw, the force in water will be:
so Fw = 80/80 = 1N.
it's the answer I think
if it helps you
mark me as brainliest plz....
F =kq1q2/r^2
k = coulomb’s constant = 1/4πεε0
ε0 = permittivity of empty space
ε = relative permittivity of the material in which the charges are immersed
r = distance between charges
for vacuum ε=1 and for air also, to a very good approximation ε=1.
For water , at room temperature ( 20C) ε = 80.
Therefore, the force between the charges q1, q2 will be 1/80 its value in air.
Hence Fw, the force in water will be:
so Fw = 80/80 = 1N.
it's the answer I think
if it helps you
mark me as brainliest plz....
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