A room with a 3 meter high ceiling has metal plate on the floor and a separate m

Question

A room with a 3 meter high ceiling has metal plate on the floor and a separate metal plate on the ceiling there is an electric field of 15,000 N/C directed upward between the plates. A ball with a mass of 6 grams and a charge of +2uC is launched horizontally from one end of the room toward the opposite wall 5 m away. The initial speed of the ball is 10m/sec and the ball is launched from a height of 2 m above the floor.

we only consider the electric force only acting on the ball. We will ignore the effects of gravitational force on the ball

a)how high above the floor will the ball hit the opposite wall?
b) with both the gravitational force and electric force acting on the ball how high above the floor will the ball hit the opposite wall? ( use g=10m/s^2)

Explanation / Answer

Hi, The force acting on a charged particle moving in an electric field is given by       F = q * E = 2 * 10^(-6) * 15000 = 0.03 N.       We also know F = m*a. Hence a = F / m = 0.03 / [6 * 10^(-3)] = 5 m/s^2 This force is acting vertically upwards and hence it imposes a displacement of the ball in +y direction. And this force will not have any effect on the horizontal displacement or horizontal velocity. The initial horizontal speed = ux = 10m/sec. Hence the time taken to travel this horizontal displacement of 5m = t                                                             = x / ux = 5 / 10 = 0.5sec. So the ball is under the influence of electric force for a time of 0.5sec. In this time, ignoring the gravitational force it will move a vertical distance of                                               y = ut + (1/2)at^2 = 0 + (0.5 * 5 * 0.5^2)                                                  = 0.625m     So the ball will strike the opposite wall at a height of h = 2 + 0.625 = 2.625m from the floor. b) If we consider the gravitational pull on the ball also, the resultant upward force acting on the ball = F' = F - Fg = 0.03 - (6 * 10^(-3) * 10) = 0.03 - 0.06 = -0.03N.    Hence the resultant acceleration = a = F' / m = -0.03 / (6 * 10^(-3)) = -5m/s^2     i.e., the resultant force will be pulling the ball downwards. In this situation, the distance the ball will be travelling vertically in 0.5 sec is given by       y = 0 - (0.5 * 5 * 0.5^2) = -0.625m. i.e., the resultant height at which the ball hits the opposite wall =
         h = 2 - 0.625 = 1.375m from the floor. Hope this helps you.

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