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sattari [20]
3 years ago
5

If the electric field between the plates has a magnitude of 1.4×105 V/m , what is the potential difference between the plates?

Physics
1 answer:
Bogdan [553]3 years ago
5 0

Answer: V = Ed

V = 107.8 Volts

The distance between plate is missing, below is a possible complete question:

A Parallel-plate Capacitor Has Plates Separated By 0.77 Mm,If the electric field between the plates has a magnitude of 1.4×105 V/m , what is the potential difference between the plates?

Explanation:

Potential difference V = electric field E × distance d

V = Ed

E = 1.4 × 10^5 V/m

d = 0.77mm = 0.77 × 10^-3 m

V = 1.4 × 0.77 × 10^(5-3) Volt

V = 1.078 × 10^2 Volts

V = 107.8 Volts

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Answer:

The puck moves a vertical height of 2.6 cm before stopping

Explanation:

As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.

So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.

Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So

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Substituting the kinetic energy of the puck for the potential energy of the spring, we have

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= 0.009971 Nm/0.38416 N

= 0.0259 m

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≅ 2.6 cm

So the puck moves a vertical height of 2.6 cm before stopping

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