Two parallel-plate capacitors have the same plate area. capacitor 1 has a plate separation twice that of capacitor 2, and the quantity of charge you place on capacitor 1 is four times the quantity you place on capacitor 2. part a how do the potential differences across each of the two capacitors compare to each other?

Explanation:

Given that:

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

.....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

From eq. (1)

For capacitor 2:

For capacitor 1:

We know, potential differences across a capacitor is given by:

..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

& for capacitor 1:

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

i think the answer would be b