An electron isreleased from rest within a horizontal electric field E hand proce

Question

An electron isreleased from rest within a horizontal electric field Ehand proceeds to accelerate to the right. (a) If it accelerates at 5x 10e9 m/s2, what is the value of Eh? (b) The lengthof the region with Eh is 10 mm. What is its finalhorizontal speed when it exits that region? (c) The electron thenenters a verticalelectric fieldEv, where the electric field strength is 3 x 106 N/Cupwards. If the electron spends 2.5 x 10e–7s in this newfield, how far does it travel to the right, how much is itdeflected from its original path, and in what direction (up ordown)?

Explanation / Answer

(a). Accleration a = 5 * 10 ^ 9 m / s ^ 2 we know a = Eq / m from this electric field   E = ma / q where m = mass of electron = 9.11 * 10 ^ -31 kg            q= charge = 1.6 * 10 ^ -19 C plug the values weget E value (b). length L = 10 mm = 10 ^ -2 m Initial speed u = 0 final speed v = ? from the relation v ^ 2 - u ^ 2 = 2aL                                       v = 2aL ] (c). the electric field strength E' = 3 x 106N/C time t ' = 2.5 * 10 ^ -7 s Accleration a ' = E ' q / m So, required deflection   S = ut ' + ( 1/ 2) at ' ^ 2

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