A pair of closely spaced parallel conducting plates, charged with equal and oppo

Question

A pair of closely spaced parallel conducting plates, charged with equal and opposite electric charges, produces a uniform electric field in the region between them. In designing a cutting-edge device that will revolutionize the electronics industry, you set up such a pair of plates with a separation of 0.923 mm between them and charge them so that the direction of the electric field in their interior region is from plate A to plate B. Your idea requires that electrons, when released from rest at one of the plates, reach the other plate at the speed of 1.71% of the speed of light. (The speed of light is c = 3.00 times 10^8 m/s.) At which plate should the electrons be released? A B Either Cannot be determined What must the strength of the electric field be?

Explanation / Answer

given that

d = 0.923 mm

v = 1.71 % of 3*10^8 = (1.71/100)*3*10^8 = 5.13*10^(6) m/s

(1)

Since the field points from A to B, and electrons are negatively charged. They will have to leave from plate B.

(2)

To find the field strength, we will apply the conservation of energy law.

The PE of the charge on plate B will equal the KE of the moving charges

PE = q*V = q*E*d

KE = 0.5*m*v2

q*E*d = 0.5*m*v2

E = 0.5*m*v2 / q*d

m = mass of electron = 9.1*10-31 kg

q = charge of electron = 1.6*10-19 C

E = (0.5)*(9.1 * 10-31)*(5.13* 106)2 / (1.6 * 10-19)*(9.23* 10-4)

E = 8.10*104 N/C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.