An experiment setup consist of two round aluminum plates(diameter= 0.4 m) that a

Question

An experiment setup consist of two round aluminum plates(diameter= 0.4 m) that are parallel to each other with a seperationof 1 cm. 1. If the plates are connected to a 48 volt supply, what isthe electric field in the space between the plates? 2. Assuming the plates are initially uncharged what is thecharge moved from one plate to the other by the batteries? a.Calculate the capacitance from the geometry, displacement and area.b. Calculate the charge from the capacitance and voltage. 3. Use the Gaussian surface method the calculate the chargefrom the Electric field. 4. What is the electrical energy stored in the plates?      An experiment setup consist of two round aluminum plates(diameter= 0.4 m) that are parallel to each other with a seperationof 1 cm. 1. If the plates are connected to a 48 volt supply, what isthe electric field in the space between the plates? 2. Assuming the plates are initially uncharged what is thecharge moved from one plate to the other by the batteries? a.Calculate the capacitance from the geometry, displacement and area.b. Calculate the charge from the capacitance and voltage. 3. Use the Gaussian surface method the calculate the chargefrom the Electric field. 4. What is the electrical energy stored in the plates?     

Explanation / Answer

            Given that the diameter of the plate is D = 0.4 m              Thesepartion between the plates is d = 1cm = 0.01m              The voltage of the battery is V = 48 V --------------------------------------------------------------------                               Then the area of the plates is A = (D/2)2      (1)  The electricfield is E = V /d                                            = 48 V / 0.01m                                            = 4800 V/m      (2) The charge is q =0AV/d                                   =-----------C          where0 = 8.85*10-12C2N-1m-2                              and q = C*V where C=0A/d     (3) From the Gauss law   E.A= q /0                                               q = EA0                                                   =---------- C      (4) The energy stored isU = (1/2)q*V                                             = -------------- J                                                            =---------- C      (4) The energy stored isU = (1/2)q*V                                             = -------------- J                                         

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