EXAMPLE 18.2 Parallel plates and conservation of energy Here we will calculate t
ID: 1644995 • Letter: E
Question
EXAMPLE 18.2 Parallel plates and conservation of energy Here we will calculate the work per unit charge on an electron moving between two potentials. A 9.0 V battery is connected across two large parallel plates that are separated by 4.5 mm of air, creating a potential difference of 9.0 V between the plates. (a) What is the electric field in the region between the plates? (b) An electron is released from rest at the negative plate. If the only force on the electron is the electric force exerted by the electric field of the plates, what is the speed of the electron as it reaches the positive plate? Video Tutor The mass of an electron is me = 9.11 × 10-31 kg. Also, U-qV, where q' =-e, the charge of an electron. Using SOLUTION in the conservation-of-energy equation gives expression to replace U SET UP Figure 18.8 shows our sketch. We tron's starting position at the negative plate and b for its final position at the positive plate. Then Vb- = +9.0 V. The electric field is direct- ed from the positive plate b toward the negative plate a (i.e., from higher potential toward lower potential), and it is uniform between the plates. use a to designate the elec- SOLVE Part (a): The expression V - Va is the potential at point b with respect to point a. This quantity (work per unit charge) is relat ed to the electric field E (force per unit charge) between the plates by bg-v, Ed, where d is the separation between the plates and Ed is the work per unit charge on a positively charged particle that moves from b to a. Thus, d 45 mm 4.5 × 10-3 m Part (b): Conservation of energy applied to points a and b at the cor- responding plates gives me-911x 10-31 g FIGURE 18.8 Our sketch for this problem.Explanation / Answer
A.
E = (Vb - Va)/d
E = 18/(4.5*10^-3)
E = 4000 V/m = 4*10^3 V/m
B.
Ka + Ua = Kb + Ub
Kb = e*(Vb - Va)
Kb = 1.6*10^-19*18 = 2.88*10^-18 J
Kb = me*vb^2/2
vb = sqrt (2*Kb/me)
vb = sqrt (2*2.88*10^-18/(9.11*10^-31)) = 2.51*10^6 m/sec
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