A20.0 μf capacitor is charged to a potential difference of 850 v. the terminals of the charged capacitor are then connected to those of an uncharged 12.0 μf capacitor. (a) compute the original charge of the system. (b) compute the final potential difference across capacitor. (c) compute the final energy of the system. (d) compute the decrease in energy when the capacitors are connected.

The properties of the capacitors can be calculated the answers are:

     a) Q₁ = 1.70 10-2 C

     b) V_f = 531 V

     c) U_f = 4.52 J

     d) ΔU = 2.71 J

Given parameters

To find

     a) The initial charge

     b) The potential difference of the system connected capacitors

     c) The final energy of the system

     d) the energy change when connecting the capacitors

A capacitor is a system formed by two separate parallel plates that serves to store electrical charge,

          Q = C V

Where Q is the stored charge, C the capacitance and V the potential difference

a) ask for the initial charge

         Q₀ = C₁ V₀

         Q₀ = 20.0 10⁻⁶  850

         Q₀ = 1.70 10⁻² C

b) The law of conservation of charge establishes that the electric charge cannot be created or destroyed, therefore the initial charge (Q₀) must be distributed between the two connected capacitors

           Q₀ =  

           C₁ V₀ = C₁  + C₂  

the Power Difference final  between the two capacitors must be the same, parallel connection

           C₁ V₀ = (C₁ + C₂)

            =

           V_f =

           V_f = 531.25 V

c) The stored energy capacitor is

          U = ½ C V²

The final energy system is

          U = ½ (C₁ + C₂)

          U = ½ (20 + 12) 10⁻⁶  531.25²

          U = 4.516 J

d) To calculate the energy change

         ΔU = U₀ -

let's look for the initial energy

         U₀ = ½ C₁ V₀²

         U₀ = ½ 20 10⁻⁶  850²

         U₀ = 7.225 J

whereby the energy change is

         ΔU = 7.225 - 4.516

         ΔU = 2.71 J

In conclusion using the properties of the capacitors we were able to calculate the answers are:

         a) Q₁ = 1.70 10-2 C

         b) V_f = 531 V

         c) U_f = 4.52 J

         d) ΔU = 2.71 J

Learn more about capacitors here:

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All four values are in 3 sig. fig.

(a)

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(b)

Sum of the final charge on the two capacitors should be the same as the sum of the initial charge. Voltage of the two capacitors should be the same. That is:

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;

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(c)

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(d)

Initial energy of the system, which is the same as the initial energy in the capacitor:

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Change in energy:

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what is the question?

what type of charge does it have? negative