Answer: 2.4×10^-3 v/m
Explanation: distance between plates of capacitor (d) =5.0×10^-3m
Potential difference between plates (v) = 12v
Force on electronic charge (f) = 3.8×10^-16 N
Strength of electric field (E) =?
The formulae that relates potential difference, eoectiic field strength and distance between plates is given as
v = Ed
By substituting the parameters, we have that
12 = E × 5.0×10^-3
E = 12/ 5.0 × 10^-3
E = 2.4×10^-3 v/m
A. Reduced greenhouse gas emissions.
Answer: A while driving a car. It is unsafe
Answer:
9.875
Explanation:
w=f×s
395=40×s
make s the subject of the formula
s=395/40
=9.875
Answer:
1.) 274.5v
2.) 206.8v
Explanation:
1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.
The potential difference and charge across EACH capacitor will be
V = Voe
Where Vo = initial voltage
e = natural logarithm = 2.718
For the first capacitor 2.50 µF,
V = Vo × 2.718
746 = Vo × 2.718
Vo = 746/2.718
Vo = 274.5v
To calculate the charge, use the below formula.
Q = CV
Q = 2.5 × 10^-6 × 274.5
Q = 6.86 × 10^-4 C
For the second capacitor 6.80 µF
V = Voe
562 = Vo × 2.718
Vo = 562/2.718
Vo = 206.77v
The charge on it will be
Q = CV
Q = 6.8 × 10^-6 × 206.77
Q = 1.41 × 10^-3 C
B.) Using the formula V = Voe again
165 = Vo × 2.718
Vo = 165 /2.718
Vo = 60.71v
Q = C × 60.71
Q = C
