A spaceship encounters a single plane of charged particles, with the charge per

Question

A spaceship encounters a single plane of charged particles, with the charge per unit area equal to ? The electric field a short distance above the plane has magnitude and is directed to the plane. (That is, in the first blank tell me the magnitude; in the second blank, tell me the direction.) 1. 2. A proton (mass 1.67x 10 "kg, charge 1.60 x 101 C) moves from point A to point B under the influence of an electrostatic force only. At point A the proton moves with a speed of 50 km/s. At point B the speed of the proton is 80 km/s. Determine the potential difference V VA

Explanation / Answer

2.

m = mass = 1.67 x 10-27 kg

q = charge = 1.6 x 10-19 C

Va = electric potential at A

Vb = electric potential at B

va = speed at A = 50,000 m/s

vb = speed at B = 80,000 m/s

using conservation of energy

q Va + (0.5) m va2 = q Vb + (0.5) m vb2

inserting the values

(1.6 x 10-19) Va + (0.5) (1.67 x 10-27) (50000)2 = (1.6 x 10-19) Vb + (0.5) (1.67 x 10-27)  (80000)2

(1.6 x 10-19) Va - (1.6 x 10-19) Vb + (0.5) (1.67 x 10-27) (50000)2 = (0.5) (1.67 x 10-27)  (80000)2

(1.6 x 10-19) (Va - Vb) + (0.5) (1.67 x 10-27) (50000)2 = (0.5) (1.67 x 10-27)  (80000)2

(Va - Vb) = 20.4 Volts

Vb - Va = - 20.4 Volts

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