please answer all. I think part a is right but I have not checked to be 100 perc
ID: 1575248 • Letter: P
Question
please answer all. I think part a is right but I have not checked to be 100 percent sure. Thanks
A water-filled parallel plate capacitor has a plate area of 2.3 cm2 and plate separation of 2.1 mm. The potential difference between its plates is held at 6.7 V. Calculate the magnitude of the electric field between its plates, the charge stored on each plate, and the charge stored on each plate after water is replaced by air. (a) the magnitude of the electric field between its plates 3190 V/m (b) the charge stored on each plate nc (c) the charge stored on each plate after water is replaced by air pC Need Help? Resadlt Viewing Saved Work Revert to Last Response -12.5 points zinPhysLS3 17.PO19. My Notes Ask Your Teache The figure below shows an electron at the origin that is released with initial speed vo 5.3 x 106 m/s at an angle 0 450 between the plates of a parallel plate capacitor of plate separation D-2.0 mm. If the potential difference between the plates is V = 160 V, calculate the closest proximity, d, of the electron to the bottom plate (in mm). .y D OExplanation / Answer
Ans:-
Given data A = 2.3cm^2,d= 2.1mm V= 6.7V
A] E =V/ d = 6.7/2.1*10^-3 = 3190V/m
B] calculate the capacitance C =0 A/d = 8.85*10^-12*2.3*10^-4/2.1*10^-3
=9.69*10^-13F
Then water increases that by 80 to 7.08 x 10-11 F.
Q = C*V = 7.08*10^-11 * 6.7 =2.96*10^-9C = 2.96nC
C]q = C*V = 9.69*10^-13 *6.7 =6.4923*10^-12C = 6.49pC
Question 2
Given data v0 =5.3*10^6m/s, =45deg, D= 2.0mm v= 160V
This problem becomes possible to solve when you realize that the electric field is uniform between the plates of the capacitor and therefore the electron feels a constant force and thus experiences a constant acceleration. This becomes a projectile motion problem except that the electric force provides the acceleration in the y–direction instead of gravity. The electric field will point down from + side to negative, and the force on the electron is
F = eE = eV / D = 1.6 × 1019C × 160V / 2 × 103 m = 1.28 × 1014 N .
The force will point up since a negative charge will always accelerate toward the direction of the electric field. The acceleration of the electron is
ay = F / m = 1.28× 1014 N / 9.1× 1031 kg = 1.4 × 10^16 m / s 2 ,
which is much larger than the acceleration due to gravity of 9.8m/s2 , justifying why we ignore gravitational force. The electron has initial speed of v0 = 5.3 × 10^6 m / s, 45deg below the horizontal as shown in figure
The y-component (vertical) is v0 y = v0 sin 45 = 3.75 × 10^6 m / s, negative since it is down.
Use vy 2 = v0 y 2 + 2ay y .
The minimum position corresponds to vy = 0 , giving y = v0 y 2 / 2ay = (3.75 × 10^6 )^2( m / s) / 2*1.4 × 10^16 m / s 2 ( ) = -5.02 × 104m . So it travels 0.5 mm below the center before reversing direction. It will reach d = 0.5 mm above the bottom plate
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