A capacitor has two parallel plates seperated by 2mm and is connected across a 50V battery. a. What is the electric field between the field? b. What is the surface charge density? c. How much charge is stored on each plate if the area is 0.1m^2. d. Calculate the capicitance e. How much energy is stored in this capacitor?

Answers

Answer 1

Answer:

a

  [tex]E =  25000 \  V/m[/tex]

b

    [tex]\sigma  =  2.2125* 10^{-7} \  C/m^2[/tex]

c

 [tex]q = 2.2125 *10^{-8} \  C  [/tex]

d

 [tex]C = 4.425 *10^{-10} \ F[/tex]

e

 [tex]U  = 5.531 *10^{-7} \  J[/tex]

Explanation:

From the question we are told that

   The distance of separation is [tex]d =  2mm =  0.002 \ m[/tex]

   The  voltage is [tex]V  = 50\  V[/tex]

   The area is  [tex]A  =0.1m^2[/tex]

Generally the  electric field is mathematically represented as

      [tex]E =  \frac{V}{d}[/tex]

=>     [tex]E =  \frac{50}{ 0.002}[/tex]

=>     [tex]E =  25000 \  V/m[/tex]

Generally the capacitance  mathematically  represented  is

       [tex]C =  \frac{\epsilon_o  * A  }{d}[/tex]

Here [tex]\episilon_o[/tex] is the permitivity of free space with value  

      [tex]\episilon_o = 8.85 * 10^{-12} C/(V⋅m   [/tex]

=>       [tex]C =  \frac{ 8.85 * 10^{-12}  * 0.1  }{0.002}[/tex]

=>      [tex]C = 4.425 *10^{-10} \ F[/tex]

Generally  the charge  is mathematically represented as

     [tex]q =  C  * V[/tex]

      [tex]q = 4.425 *10^{-10}   *50 [/tex]

=>    [tex]q = 2.2125 *10^{-8} \  C  [/tex]

Generally the charge density is  mathematically represented as

       [tex]\sigma  =  \frac{q}{A}[/tex]

=>   [tex]\sigma  =  \frac{ 2.2125 *10^{-8}}{0.1}[/tex]

=>   [tex]\sigma  =  2.2125* 10^{-7} \  C/m^2[/tex]

Generally the energy stored in this capacitor is mathematically

    [tex]U  = \frac{1}{2} *  CV^2[/tex]

=>  [tex]U  = \frac{1}{2} *  (4.425 *10^{-10} ) * (50)^2[/tex]

=>   [tex]U  = 5.531 *10^{-7} \  J[/tex]


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i got it right on ed 2022

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Answers

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