Delta K = -Delta U Delta U = q Delta V Delta V = Eh K = 1/2 mv^2 A negative ion

Question

Delta K = -Delta U Delta U = q Delta V Delta V = Eh K = 1/2 mv^2 A negative ion (q = -3.20 times 10^-19 C), with mass 3.68 times 10^-26 kg, is accelerated across a uniform electric field between two oppositely-charged parallel plates. The distance from o plate to another is 16.5 cm. The magnitude of the electric field is 1, 100 N/C. (a.) What is the direction of the electric field between the plates? (b.) Which plate (right/left) is at a higher potential? (c.) What is the potential difference between the plates? (d.) If the ion begins at rest, how much kinetic energy does it gain by the time it reaches the opposite plate?

Explanation / Answer

(A) ion is accelerated so force on ion will be to the right.

and F = q E

q is naegative hence E will be to the left .

(b) field (E) is always from higher potential to lower potential.
  
Right plate is at higher potential.

(c) V = E.d = (1100)(0.165)

= 181.5 Volt

(d) KE gain = q deltaV

= (3.20 x 10^-19) (181.5)

= 5.808 x 10^-17 J

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