Question

A parallel-plate air capacitor is made from two plates 0.200{\rm m} square, spaced 0.800 {\rm cm} apart. It is connected to a 120 {\rm V} battery. If the plates are pulled apart to aseparation of 1.60 {\rm cm}, suppose the battery remains connected while theplates are pulled apart.
What is the capacitance?

C =2.21×10−11{\rm F}


Part B

What is the charge on each plate?

Enter your answer as two numbers,separated with a comma.

Part C

What is the electric field between theplates?

E =7500 {\rm V/m}

Part D

What is the energy stored in thecapacitor?

U =1.59×10−7{\rm J}

Answer 1

Answer:

A)  C = 4.425 10⁻¹³ F,  B) Q = 6.31 10⁻¹¹ C,  C)  E = 8.9 10⁴ N / C,  

D)  u_{E} = 3.19 10⁻⁹ J

Explanation:

Part A. Capacitance is

           C = ε₀ A / d

in this case distance is d = 0.800 cm = 0.800 10⁻² m and the area of ​​the plates is x = 0.200 cm= 0.2 10⁻² m, all the quantities must be in the SI system for the result to be in Farads

           A = x²

           A = (0.2 10⁻²)²

           A = 4 10⁻⁶ m

let's calculate

          C = 8.85 10⁻¹² 4 10⁻⁶ / 0.8 10⁻²

          C = 44.25 10⁻¹⁴ F

          C = 4.425 10⁻¹³ F

Part B. The charge on each plate is

          Q = C ΔV

           Q = 4.425 10⁻¹³ 120

           Q = 6.31 10⁻¹¹ C

Part C. the electric field of a plate is

           E = σ / 2ε₀

where the charge density is

            σ = Q / A

we substitute

             E =$\frac{Q}{ 2\epsilon_o A}$

       

let's calculate

             E = $\frac{6.31^{-11} }{ 2 \ 8.85^{-12} \ 4^{-6} }$

             E = 8.9 10⁴ N / C

Part D. stored energy

              $u_{E}$ = ½ C V²

               u_{E} = ½ 4.425 10⁻¹³ 120²

               u_{E} = 3.19 10⁻⁹ J