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xz_007 [3.2K]
2 years ago
9

A parallel plate capacitor fully charged to voltage V is connected to the battery (the voltage on the plates remains fixed). If

the plates are pulled away from each other, how would it affect the charge on the capacitor
Physics
1 answer:
Papessa [141]2 years ago
7 0

Charge will decreases.

A parallel plate capacitor when it is fully charged to voltage V is given as:

                    C = Q/V

The capacitance of parallel plate capacitor with two plates of Area A separated by distance d and no dielectric material between plates is

                    C = ε₀ A /d

since from above equation it shows C is proportional to Q and also C is inversely proportional to distance d.

So, ATQ when d increases C will decrease which in result decreases charge on the capacitor.

Thus,  Charge will decrease.

Learn more about capacitance here:

     brainly.com/question/17115454

          #SPJ4

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single goose sounds a loud warning when an intruder enters the farmyard. Some distance from the goose, you measure the sound lev
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Answer:

The sound level of the 26 geese is  Z_{26}= 96.15 dB

Explanation:

From the question we are told that

    The  sound level is Z_1 =  81.0 \ dB

     The number of geese is N = 26

Generally the intensity level of sound is mathematically represented as

        The intensity of sound level in dB  for one  goose is mathematically represented as

                       Z_1 = 10 log [\frac{I}{I_O} ]

Where I_o is the  threshold level of intensity with value  I_o = 1*10^{-12} \  W/m^2

            I is the intensity for one goose in W/m^2

For 26 geese the intensity would be  

          I_{26} = 26 * I

   Then  the intensity of 26 geese in dB is  

              Z_{26} = 10 log[\frac{26 I }{I_o} ]

               Z_{26} = 10 log (\ \ 26 *  [\frac{ I }{I_o} ]\ \ )

               Z_{26} = 10 log (\ \ 26  \ \ ) *   (\ \  10 log [\frac{ I }{I_o} ]\ \ )

 From the law of logarithm we have that

              Z_{26} = 10 log 26 +  10 log [\frac{I}{I_0} ]

                    = 14.15 + 82

                    Z_{26}= 96.15 dB

               

               

           

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Sphere 1 with radius R_1 has positive charge q, Sphere 2 with radius 4.50 R_1 is far from sphere 1 and initially uncharged. The
tia_tia [17]

Answer:

Explanation:

capacitance of sphere 2 will be 4.5 times sphere 1

a ) when spheres are in contact they will have same potential finally . So

V_1 / V_2 = 1

b )

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charge on sphere1 = q  x 1 / ( 1 + 4.5 )

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