The electric field between the two plates shown below has a magnitude of 370 N/C

Question

The electric field between the two plates shown below has a magnitude of 370 N/C. The plates are separated by a distance d of 1.9 cm. What is the magnitude of the potential difference VBA between the two plates (in V)? What is the speed of a proton (in m/s) released from rest very close to plate A just before it hits plate B? (Enter the magnitudes.)

(Assume that the system is in a vacuum and the surface area of the plates is much larger than the distance between the plates. Note

VBA = Vf Vi = VB VA.)

Explanation / Answer

Given,

E = 370 N/C ; d = 1.9 cm = 0.019 m ;

We know that

E = V/d => V = Ed

Vb - Va = 370 x 0.019 = 7.03 V

Hence, Vb - Va = 7.03 V

for speed,

W = q V ; W = 0.5 m v^2

q V = 1/2 m v^2

v = sqrt (2 q V/m)

v = sqrt (2 x 1.6 x 10^-19 x 7.03/(9.1 x 10^-31) = 1.57 x 10^6 m/s

Hence, v = 1.57 x 10^6 m/s

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