Answer:
The capacitor having less distance of separation has a stronger electric field.
Explanation:
The capacitors are identical and only difference between them is that one has twice the plate separation of the other. Therefore, capacitance of the given capacitors C1 and C2 is,
C1= Aε/d and C2=Aε/2d
The charges Q1 and Q2 on the capacitors of capacitance C1 and C2 respectively, is then given by the equation,
Q1=VC1
Q1=VAε/d
Q2=VC2
Q2=VAε/2d
Therefore, the surface charge density σ1 and σ2 for the capacitors is,
σ1=Q1/A
σ1=VAε/(d*A)
σ1=Vε/d
Similarly,
σ2=Q2/A
σ2=Vε/2d
The electric field between the plates is directly proportional to the surface charge density. And so electric field is inversely proportional to the distance of separation. Therefore the capacitor whose distance of separation is less has a stronger electric field.
Check out similar questions.
brainly.com/question/14744776
#SPJ10
<span>The answer would be b because b has a set of arrows represents the induced magnetic force</span>
Answer:
yes i relate mass but not acceleration
Dropping two objects and seeing them hit the ground at the same time
Answer:
Part A:
Part B:
Explanation:
Part A:
To calculate the number of free electrons n we use the following formula::
n=1.5N-Au
Where N-Au is number of gold atoms per cubic meter
So:
Part B:
n is calculated above which is 8.85*10^{28}m^{-3}
Charge on electron=1.602*10^{-19}
Elec- Conductivity= 4.3*10^{7}
