An electron is placed midway between two parallel conducting plates that are spaced 3.0 mm apart. The plates are attached to the terminals of a 12.0 V battery.
a) What is the direction and magnitude of the electric field?
b) How much work would be done by moving the electron from the midpoint to the surface of the (-) plate?
c) If the electron were released from a point midway between the plates, what would be its velocity when it struck the (+) plate?
d) If the potential were doubled, how much faster would the velocity be?
Any help would be greatly appreciated. Easy 10 points for a decent explanation.
a) What is the direction and magnitude of the electric field?
b) How much work would be done by moving the electron from the midpoint to the surface of the (-) plate?
c) If the electron were released from a point midway between the plates, what would be its velocity when it struck the (+) plate?
d) If the potential were doubled, how much faster would the velocity be?
Any help would be greatly appreciated. Easy 10 points for a decent explanation.
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(a) By E = V/d = 12/(3 x 10^-3) = 4000 V/m or N/C
(b) By W = Fd
=>W = (eE) x d
=>W = 1.6 x 10^-19 x 4000 x 1.5 x 10^-3
=>W = 9.6 x 10^-19 J
(c) By the work energy relation:-
=>W = ∆KE
=>9.6 x 10^-19 = 1/2 x 9.1 x 10^-31 x v^2
=>v = √[2.11 x 10^12]
=>v = 1.45 x 10^6 m/s
(d) By the work energy relation:-
=>W = ∆KE
=>2 x 9.6 x 10^-19 = 1/2 x 9.1 x 10^-31 x v^2
=>v = √[4.22 x 10^12]
=>v = 2.05 x 10^6 m/s
(b) By W = Fd
=>W = (eE) x d
=>W = 1.6 x 10^-19 x 4000 x 1.5 x 10^-3
=>W = 9.6 x 10^-19 J
(c) By the work energy relation:-
=>W = ∆KE
=>9.6 x 10^-19 = 1/2 x 9.1 x 10^-31 x v^2
=>v = √[2.11 x 10^12]
=>v = 1.45 x 10^6 m/s
(d) By the work energy relation:-
=>W = ∆KE
=>2 x 9.6 x 10^-19 = 1/2 x 9.1 x 10^-31 x v^2
=>v = √[4.22 x 10^12]
=>v = 2.05 x 10^6 m/s