A uniform field is established between two horizontal plates by connecting them

Question

A uniform field is established between two horizontal plates by connecting them to a 100 volt battery, with the upper plate at the higher potential. The plates are in a vacuum and are 6 cm apart. A plastic sphere with a mass of 3.0 x 10^-12 kg and a charge of -0.035 microcoulombs is placed on the lower plate.
(a) What is the electric field between the plates?
(b) What is the electric potential energy of the sphere with respect to the upper plate?
(c) If the sphere is released, what will be its kinetic energy when it hits the upper plate?
(d) How fast will it be going?
(e) How great a force propelled it upward?

Explanation / Answer

Given Voltage V   = 100 V distance between the plates   d =   0.06 m mass   of the sphere m = 3.0*10^-12 kg charge q     = - 0.035*10^-6 C ------------------------------------------- (a) Electric field between the plates   E = V/d   = 100 / 0.06   =  1666.67 N/C (b) The electric potensial energy of the sphere between the plates U =qV                            U = ( 0.035*10^-6 C)(100)   = 3.5*10^-6 j (c) The force acting on the sphere According to newton's second law F = ma   = qE Acceleration of the sphere a = qE / m    = (0.035*10^-6 C) ( 1666.67 N/C) / (3.0*10^-12 kg) a   = 1.944*10^7 m/s (d) According to kinematic relation v^2 - u^2 = 2ad initial velocity u = 0.0 m/s , d =   0.06 m final velocity of the sphere when it hit the upper plate v = sqrt ( 2*a*d) = sqrt  (2*1.944*10^7 m/s^2 ) (0.06))   = 1.53*10^3 m/s (c) the kinetic energy of the sphere when it hits the upper plate K.E = (1/2) mv^2 = (1/2) (3*10^-12 kg ) (1.53*10^3)^2 = 3.5*10^-6 j (e) The upward force acting on the sphere F = qE   = (0.035*10^-6 C) (1666.67 N/C) = 5.83*10^-5 N

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