At a given instant, a 2.2 A current flows in the wires connected to a parallel-plate capacitor. What is the rate at which the el
ectric field is changing between the plates if the square plates are 2.0 cm on a side? Express your answer using two significant figures
1 answer:
Answer:
Check attachment for better understanding
Explanation:
Given that,
Current in wire I =2.2A
Capacitor plate dimension is 2cm by 2cm
s=2cm=2/100 = 0.02m
Rate at which electric field Is changing dE/dt?
The current in the wires must also be the displacement current in the capacitor. We find the rate at which the electric field is changing from
ID = ε0•A•dE/dt
Where ε0 is a constant
ε0= 8.85×10^-12C²/Nm²
Area of the square plate is
A =s² =0.02² = 0.0004m²
Then,
Make dE/dt the subject of formula
dE/dt = ID/ε0A
dE/dt = 2.2 / (8.85×10^-12 ×4×10^-4)
dE/dt = 6.215×10^14 V/m-s
Or
dE/dt = 6.215×10^14 N/C.s
The rate at which the electric field is changing between the plates is 6.215×10^14 N/C.s
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Answer:
The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.
Explanation:
Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:
and since the electric field E in between parallel plates separated a distance d and under a potential difference , is given by:
then :
We want to find when the particle reaches velocity zero via kinematics:
We replace this time (t) in the kinematic equation for the particle displacement:
Replacing the values with the information given, converting the distance d into meters (0.01 m), using , and the electron's kinetic energy:
we get:
Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:
0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]