8) A charged particle with charge +5.0 C is initially at rest. It is accelerated

Question

8) A charged particle with charge +5.0 C is initially at rest. It is accelerated through a potential 8) difference of 500 V. What is the kinetic energy of this charged particle? A) 2.5 10-3J B) 0J C) 2500J D) 2500 J/m E) 2.5 × 10-3 J/m 9) The potential difference between the plates of a parallel plate capacitor with the plate separation of 9) 6 cm is 60 V. What is the electric field between the plates of this capacitor? A) 2000 V/m B) 500 V/m C) 60 V/m D) 1000 V/m E) 3600 V/m 10) A parallel plate capacitor with plate separation of 4.0 cm has a plate area of 4.0x 10-2 m2, What is 10) the capacitance of this capacitor with air between these plates? A) 8.9 × 10-12 F B) 8.9 × 10-14 F C) 8.9 x 10-11 F D) 8.9 x 10-15 F E) 8.9 x 10-13 F

Explanation / Answer

8) given cahrge q = 5 mico coulombs, potential difference delta v = 500 V
now the change in potential energy D U = DV*q = 500*5*10^-6 = 2.50 mJ

change in kinetic energy of the particle which is accelerated thorugh a potential differene is

equal to the change in potential energy

   change in kinetic energy = - DU = - 2.5 mJ

Answer is option E 2.5 *10^-3 J

9)

   parallel plate capacitor

   separation of plates d = 6 cm, potential difference V = 60 V

   electric field between the plates is E = v/d

   E = sigma/ epsilon not = q/A *epsilon not = cv/A*epsilon not = epsilon not *A *v/d*A *epsilon not = v/d

   E = 60/6*10^-2 = 1000 V/m

       option D is answer

10) capacitance of the parallel plate capacitor C = epsilon not *A/d
                       = 8.9*10^-12*4*10^-2/4*10^-2 F

   Answer is option A

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