Two parallel-plate capacitors, 2.6 μf each, are connected in parallel to a 28 v battery. one of the capacitors is then squeezed so that its plate separation is halved. because of the squeezing, (a) how much additional charge is transferred to the capacitors by the battery and (b) what is the increase in the total charge stored on the capacitors?

a.

b.

Explanation:

We are given that

Capacitance of one capacitor=

Capacitance of second capacitor=

Potential difference of battery=28 v

a.When one of the capacitor is squeezed so that plate separation is halved then we have to find the amount of additional charge is transferred to the capacitor by the battery.

Initial capacitance,

If the separation is halved , then the capacitance will be  doubled

If the capacitance will be doubled then the charge will be doubled also.

Initial charge=

Final charge on the capacitor

Additional charge transmitted,

b.We have to find the increase in the total charge stored on the capacitors

Initial total charge is given by

Final total charge on the capacitor

Increase in the charge=

  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

the answer is b

explanation: