Two 1 m times 1 m conducting plates are spaced 10 cm and oriented horizontally(e

Question

Two 1 m times 1 m conducting plates are spaced 10 cm and oriented horizontally(each plane is parallel to the floor). A 15 V battery is connected to the plates with the positive terminal connected to the lower plate. What is the electric field (magnitude and direction) between the plates? Explain. What is the total charge on the upper plate (sign and magnitude)? Explain. What is the capacitance of the two plates? Explain. A proton is released midway between the plates (5 cm from each plate in the vertical direction & 0.5 m horizontally from the front and left edge of each plate). A. What plate does the proton hit? Explain. B. What force (magnitude and direction) acts on the proton immediately after it is released? Explain. C. What is the potential midway between the two planes? Explain. D. What is the proton speed when it hits the plate? Explain. E. How long does it take the proton to hit the plate? Explain An electron enters the gap between the plates from the left midway between the plates (5 cm vertically from each plate & centered on the plate width), traveling horizontally at 500 m/s. A. Which plate does the electron get closer to? Explain. B. Does the electron leave the gap between the two plates before hitting one or the other plate? Explain.

Explanation / Answer

1.After connecting the conducting plates to the battery one is positively charged and other is negative charged.Since lines of force directed away from positive charge and towars a negative charge the direction of electric field will be from positive plate to negative plate.

Magnitude of electric field-

Area of the plate=1 m* 1 m=1m2

Capacitance "C" = q/V=0A/D (D=Distance between plate)

=8.85*10-12F/m*1m/0.1m=8.85*10-11F/m2

Now,Charge "q"=C*V=8.85*10-11F/m2*15V=132.75*10-11 C

Electric Field Fe=9*109*132.75*10-11/0.1=119.475 N/C Along negative plate.

2. The total charge on the upper plate is 132.75*10-11 C and it is positive sign.

3. The capacitance of the two plate are 8.85*10-11F.

4.A. Since the proton is positively charged and like charges repel each other and unlike charges attract each other. Therefore proton is attracted by negative plate.

B. Attractive force produced by the negative plate is acts upon the proton.

C. The potential midway between the plate is same as the potential between the plate i.e. 15 V.

5.A. Since electron is negatively charged and it attracted by the positive charge.So it can get closer to the positive charged plate.

B. If the velocity of the electron is very high then it may have pass the gap between two plate without hitting it.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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