Oppositely charged parallel plates are separately by 4.61 mm. A potential differ

Question

Oppositely charged parallel plates are separately by 4.61 mm. A potential difference of 600 V exists between the plates. What is the magnitude of the electric field between the plates? What is the magnitude of the force on an electron between the plates? How much work must be done on the electron to move it to the negative plate if it is initially positions 2.30 mm from the positive plate? In Rutherford's famous scattering experiments that led to the planetary model of the atom, alpha parties (having charges of +2e and masses of 6 64 times 10^-27 kg) were fire toward a gold nucleus with charge +79e. An alpha particle, initially very far from the gold nucleus, is fired at 1.4 times 10^7 m/s directly toward the nucleus, as is the figure below How close does the alpha particle get to the gold nucleus before turning around? Assume the gold nucleus remains stationary.

Explanation / Answer

a) Electric filed intensity = Potential difference / Distance
Electric filed intensity = 600 / (4.61x10^-3) =130151.84 V/m

b) Electric filed intenisty = Force / Charge
Charge of an electron = 1.6x10^-19 C
Electric filed intenisty = 130151.84 V/m
Force = 130151.84 x 1.6x10^-19 = 2.08 x10^-14 N

c) Change in potential energy = Force x Distance
Change in potential energy = 2.08 x10^-14 x (2.30x10^-3 )
Change in potential energy = 4.784 x10^-17 J

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