Answer:
A) 26V
Explanation:
(a) the potential difference between the plates
Initial capacitance can be calculated using below expresion
C1= A ε0/ d1
Where d1= distance between = 2.70 mm= 2.70× 10^-3 m
ε0= permittivity of space= 8.85× 10^-12 Fm^-1
A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2
If we substitute the values we
C1= A ε0/ d1
=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3
C1=2.589 ×10^-12 F= 2.59 pF
Initial charge can be determined using below expresion
q1= C1 × V1
V1=2.589 ×10^-12 F
V1= voltage=7.90 V
If we substitute we have
q1= 2.589 ×10^-12 × 7.90
q1= 20.45×10^-12C
20.45 pC
Final capacitance can be calculated as
C2= A ε0/ d2
d2=8.80 mm= /8.80× 10^-3
7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3
C1=0.794 ×10^-12 F= 0.794 pF
Final charge= initial charge
q2=q1 (since the battery is disconnected)
q2=q1= 20.45 pC
Final potential difference
V2= q/C2
= 20.45/0.794
= 26V
Answer:
8.1 x 10^13 electrons passed through the accelerator over 1.8 hours.
Explanation:
The total charge accumulated in 1.8 hours will be:
Total Charge = I x t = (-2.0 nC/s)(1.8 hrs)(3600 s/ 1 hr)
Total Charge = - 12960 nC = - 12.96 x 10^(-6) C
Since, the charge on one electron is e = - 1.6 x 10^(-19) C
Therefore, no. of electrons will be:
No. of electrons = Total Charge/Charge on one electron
No. of electrons = [- 12.96 x 10^(-6) C]/[- 1.6 x 10^(-19) C]
<u>No. of electrons = 8.1 x 10^13 electrons</u>
Answer:
5 is the tripoid stand
Thanks have a bangtastic day
The hull type that is best for use on ponds, small lakes and calm rivers is Flat Bottom Hull.
A flat bottomed boat is a boat with a flat bottomed, two-chined hull, which allows it to be used in shallow bodies of water, such as rivers, because it is less likely to ground. The flat hull also makes the boat more stable in calm water.
<span>C. It is the difference in electrical potential energy between two places in an electric field.</span>
