Two 1m X 1m conducting plates are spaced 10cm apart and oriented horizontally (

Question

Two 1m X 1m conducting plates are spaced 10cm apart and oriented horizontally ( each plate is parallel to the floor). A 15V battery is connected to the plates with the positive terminal connected to the lower plate.

1) What is the electric field (magnitude and direction) between the plates? Explain

2) What is the total charge on the upper plate? (sign and magnitude) Explain.

3) What is the capacitance of the two plates?

A proton is released midway between the plates (5cm from each plate in the vertical direction and .5 horizontally from the front and left edge of each plate)

1) What plate does the proton hit? Explain

2) What force (magnitude and direction) acts on the proton immediately after it is released? Explain.

3) What is the potential midway between the two plates?

4) what is the proton speed when it hits the plate? Explain.

5) How long does it take for the proton to hit the plate?

An electron enters the gap between the plates from the left midway between the plates (5cm vertically from each plate and centered on the plate width). teavelling horizontally at 500 m/s

1) Which plate does the electron get closer to?

2) Does the electron leave the gap between the two plates before hitting one or the other plate? Explain

Explanation / Answer

1) E = V / d = 15 Volt / (0.10 m) = 150 V/m

2) E = sigma/ e0 = Q / (e0 A)

150 = (Q) / (8.854 x 10^-12 x 1 x 1)

Q = 1.328 x 10^-9 C

upper plate will be negatively vharge.

Q = - 1.328 x 10^-9 C

3) Q = C V

C = 1.328 x 10^-9 / 15 = 8.854 x 10^-11 F

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