Answer:11.1
Explanation:
Three significant figures
The electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.
To determine σ:
σ = Q/A
Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:
σ = Q/d²
Make this substitution in the equation for E:
E = Q/(2ε₀d²)
We see that E is inversely proportional to the square of d:
E ∝ 1/d²
The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude:

A) one ... same current passes through each cpt
Answer:
i think it is c if not im sorry if im wrong
Explanation:
It will take 267 milliseconds for a sample of radon-218 to decay from 99 grams to 0. 50 grams.
We know that half life of a first order reaction is given by: 
where k = rate of reaction
Given half life = 35 milliseconds
So from this we get k = 0.0198
Now we know that rate of first order reaction is given by: 
where t= time
R'= initial amount = 99 g
R= final amount= 0.50 g
k= rate of reaction = 0.0198
Putting values of these in above equation we get t=267 milliseconds.
i.e. It will take 267 milliseconds for a sample of radon-218 to decay from 99 grams to 0. 50 grams.
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