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3 years ago

7

A 4.15-volt battery is connected across a parallel-plate capacitor. Illuminating the plates with ultraviolet light causes electr

ons to be emitted from the plates with a speed of 1.76 × 106 m/s. (a) Suppose electrons are emitted near the center of the negative plate and travel perpendicular to that plate toward the opposite plate. Find the speed of the electrons when they reach the positive plate. (b) Suppose instead that electrons are emitted perpendicular to the positive plate. Find their speed when they reach the negative plate.

1 answer:

FromTheMoon [43]3 years ago

6 0

Answer:

a ) 2.13 X 10⁶ m/s .

b ) 1.28 X 10⁶ m/s

Explanation:

When electrons are repelled by negative plates and attracted by positive plates , it will increase their kinetic energy.

Increase in their energy = 4.15 eV

= 4.5 X 1.6 X 10⁻¹⁹ J

= 6.64 x 10⁻¹⁹ J

Initial kinetic energy

= 1/2 mv²

= 1/2 x 9.1 x 10⁻³¹ x ( 1.76 x 10⁶)²

= 14.09 X 10⁻¹⁹ J

Total energy

= 6.64 x 10⁻¹⁹+14.09 X 10⁻¹⁹

= 20.73 x 10⁻¹⁹

If V be the increased velocity

1/2 m V² = 20.73 X 10⁻¹⁹

.5 X 9.1 X 10⁻³¹ V² = 20.73 X 10⁻¹⁹

V = 2.13 X 10⁶ m/s .

b ) When electrons are released from positive plate , their speed are reduced because of attraction between electrons and positively charge plates.

Initial kinetic energy

= 14.09 x 10⁻¹⁹ J (see above )

reduction in kinetic energy

= 6.64 x 10⁻¹⁹ J ( See above )

Total energy with electron -

= 14.09 x 10⁻¹⁹ - 6.64 x 10⁻¹⁹

= 7. 45 x 10⁻¹⁹ J .

If V be the energy of electrons reaching the negative plate,

1/2 m V² =7. 45 x 10⁻¹⁹

V = 1.28 X 10⁶ ms⁻¹

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