1) Now we also know that the charge on a pair of parallel plates isproportional
ID: 1754451 • Letter: 1
Question
1) Now we also know that the charge on a pair of parallel plates isproportional to the potential difference V, namelyq=CV.
We also know that the capacitance of two parallel plates is givenby
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) E
B) C
C) 0
D) 0A/2
2) So now let's say that you find a slope of8.376×10-14 for square plates of width 13.0 cm.What do you get for the value of the permittivity of free space?(Ignore units for now.)
3) If the uncertanty in the slope is 7.036×10-15,is the measured value of 0 withinexperimental uncertainty of the accepted value? (Enter Y for yesand N for no)
Explanation / Answer
we know that force F =(1/2)0AE2 If we draw graph between the F and E2 we will get a straight line passing through origin. Then slope of the graph is (1/2)0A Given that slope of the graph is8.376×10-14 Width of the plates is s = 13 cm = 0.13 m Then area of the plates is A =s2=(0.13)2 m2 Then (1/2)0A = 8.376×10-14 0A= 2* 8.376×10-14 0= 2* 8.376×10-14 /A= 2* 8.376×10-14 /(0.13)2 C2/Nm2 =9.91*10-12 C2/Nm2Then uncertainity in slope slope =(1/2)0*A 7.036×10-15 = (1/2)0*(0.13)2 m2 Then 0 =0.833*10-12C2/Nm2 As the experimental value of 0 =8.854*10-12C2/Nm2 Then the calculated value is 9.91*10-12C2/Nm2
then Differcence between the experimental and actual valuesis 9.91*10-12C2/Nm2 -8.854*10-12C2/Nm2 =1.056C2/Nm2 Thus measured value of 0 withinexperimental uncertainty of the accepted value is false.
0A= 2* 8.376×10-14 0= 2* 8.376×10-14 /A= 2* 8.376×10-14 /(0.13)2 C2/Nm2 =9.91*10-12 C2/Nm2
Then uncertainity in slope slope =(1/2)0*A 7.036×10-15 = (1/2)0*(0.13)2 m2 Then 0 =0.833*10-12C2/Nm2 As the experimental value of 0 =8.854*10-12C2/Nm2 Then the calculated value is 9.91*10-12C2/Nm2
then Differcence between the experimental and actual valuesis 9.91*10-12C2/Nm2 -8.854*10-12C2/Nm2 =1.056C2/Nm2 Thus measured value of 0 withinexperimental uncertainty of the accepted value is false.
0= 2* 8.376×10-14 /A= 2* 8.376×10-14 /(0.13)2 C2/Nm2 =9.91*10-12 C2/Nm2
Then uncertainity in slope slope =(1/2)0*A 7.036×10-15 = (1/2)0*(0.13)2 m2 Then 0 =0.833*10-12C2/Nm2 As the experimental value of 0 =8.854*10-12C2/Nm2 Then the calculated value is 9.91*10-12C2/Nm2
then Differcence between the experimental and actual valuesis 9.91*10-12C2/Nm2 -8.854*10-12C2/Nm2 =1.056C2/Nm2 Thus measured value of 0 withinexperimental uncertainty of the accepted value is false.
Then the calculated value is 9.91*10-12C2/Nm2
then Differcence between the experimental and actual valuesis 9.91*10-12C2/Nm2 -8.854*10-12C2/Nm2 =1.056C2/Nm2 Thus measured value of 0 withinexperimental uncertainty of the accepted value is false.
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