we also know that the charge on a pair of parallel plates isproportional to the

Question

we also know that the charge on a pair of parallel plates isproportional to the potential difference V, namely q=CV. We also know that thecapacitance of two parallel plates is given by
C=0A/d,
where 0=8.85×10-12 isthe permittivity of free space and d is the separationbetween the plates. Combine these equations and to get the force asa function of the potential. So now if you were to make a graph offorce (F) vs.V2/d2, the slope of thegraph would be

A) 0
B) 0A/2
C) E
D) C


The electric field near the surface of a uniformly chargedconducting plane can be determined by Gauss's law (look it up). Ifthe plane is a square of sides 16.5 cm and carries a charge of 35.0nC (that's nanocoulombs), what is the magnitude of the electricfield near the surface?




Now let's say that a plane with a charge of equal magnitude butopposite sign is brought near the first. What is the force betweenthe two planes?




So now let's say that you find a slope of1.120×10-13 for square plates of width 16.5 cm.What do you get for the value of the permittivity of free space?(Ignore units for now.)







If the uncertanty in the slope is 3.921×10-15, isthe measured value of 0 withinexperimental uncertainty of the accepted value? (Enter Y for yesand N for no)

Explanation / Answer

Given : We know that   q = c v capacitance of two parallel plates isC=0A/d,
where 0=8.85×10-12 isthe permittivity of free space and d is the separationbetween the plates.    a graph of force (F) vs.V2/d2, the slope of thegraph would be B)0A/2 I hope it helps you

I hope it helps you

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