Integrated Concepts Figure 18.57 shows an electron passing between two charged metal plates that create an 100 N/C vertical electric field perpendicular to the electron’s original horizontal velocity. (These can be used to change the electron’s direction, such as in an oscilloscope.) The initial speed of the electron is 3.00×10? m/s , and the horizontal distance it travels in the uniform field is 4.00 cm. (a) What is its vertical deflection? (b) What is the vertical component of its final velocity? (c) At what angle does it exit? Neglect any edge effects.

Solution : In the above figure shows an electron passing between two charged metal plates that create an 100 N/C vertical electric field perpendicular to the electron’s original horizontal velocity. The initial speed of the electron is 3.00×10 m/s and The horizontal distance it travels in the uniform field is 4.00 cm. Step 1 Refer to figure 18.57 provided in the textbook, (a) What is its vertical deflection Before the above question we want to know the vertical deflection Vertical deflection: The vertical deflection at a point on the earth is a measure of how far the direction of the local gravity field has been shifted by local anomalies such as nearby mountains. Then from the above vertical deflection have been mentioned formula calculate time in which the electron travels a horizontal displacement, L = v t ox Here, L is the horizontal distance travelled by electron v ox is the initial horizontal velocity of electron, and t is time taken by electron before coming to rest. Therefore, rewrite the above equation is, t = L v ox Then the magnitude of electrostatic force acting in vertical direction is, F = qE where, q E is the electric field between the plates.