Two horizontal metal plates, each 10.0 cm square, are aligned 1.00 cm apart with

Question

Two horizontal metal plates, each 10.0 cm square, are aligned 1.00 cm apart with one above the other. They are given equal-magnitude charges of opposite sign so that a uniform downward electric field of 1.82 times 103 N/C exists in the region between them. A particle of mass 2.00 times 10-16 kg and with a positive charge of 1.04 times 10-6 C leaves the center of the bottom negative plate with an initial speed of 1.04 times 105 m/s at an angle of 37.0 degree above the horizontal. Where does it strike, relative to its starting point? (Enter the horizontal distance from the initial position.)

Explanation / Answer

The total time of flight T = 2 u / a

= 2u * sin 37 / a

The horizontal distance travelled in this time interval is given by

x = (u cos 37) T

= (u cos 37) (2u * sin 37 / a)

= u^2 ( 2 sin 37 cos 37) / [qE/m]

= m U^2 sin 74 / qE

= (2 * 10-16 * (1.04 * 10^5)^2 * sin 74) / (1.04 *10-6 * 1.82 * 10^3)

= 1.0986 mm

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