物理高手快來~~ - Ans.fyi 參考答案

物理高手快來~~

[匿名]

2009-03-05 7:09 am

1.Find the ratio of the electric force between a proton andd an electron to the gravitational force between the two. Why doesn't it matter that you aren't told the distance between them?
2.Earth carries a net charge of -4.3*10^5 C. The force due to this charge is the same as if it were concentrated at Earth's center.How much charge would you have to place on a 1.0-g mass in order for the electric and gravitational forces on it to balance?
3.An electron placed in an electric field experiences an electric force of 6.1-10^-10N.What is the field strength?
4.What is the magnitude of the force on a 2.0-mC charge in a 100-N/C electric field?
5.The electric field 22 cm from a long wire carrying a uniform line charge density is 1.9kN/C.What will be the field strength 38 cm from the wire?
6.In his famous 1909 experiment that demonstrated quantization of electric charge, R. A. Millikan suspended small oil drops in an electric field . With a field strength of 20MN/C,What mass drop can be suspended when the drop carries a net charge of 10 elementary charges?
7.How strong an electric field is needed to accelerate electrons in a TVtube from rest to one-tenth the speed of light in a distance of 5.0 cm?

回答 (1)

Prof. Physics

2009-03-05 4:23 pm

✔ 最佳答案

1. Charge of proton, Q = 1.6 X 10^-19 C

Charge of electron, q = -1.6 X 10^-19 C

Mass of proton, M = 1.66 X 10^-27 kg

Mass of electron, m = 9.11 X 10^-31 kg

Both electric force and gravitational force obeys inverse square law

Required ratio

= (Qq/4πεo r^2 ) / (GMm / r^2)

= Qq / 4πεo GMm

= (1.6 X 10^-19)(1.6 X 10^-19) / [4π(8.85 X 10^-12)(6.67 X 10^-11)(1.66 X 10^-27)(9.11 X 10^-31)]

= 2.28 X 10^39 (only magnitude is considered)

From the equation, we see the required ratio is independent of the distance between them.

2. By GMm / r^2 = Qq/4πεo r^2

(6.67 X 10^-11)(5.98 X 10^24)(0.001) = (-4.3 X 10^5)q / 4π(8.85 X 10^-12)

Required charge, q = -1.03 X 10^-4 C (negative charge)

3. By F = qE

6.1 X 10^-10 = 1.6 X 10^-19 E

Field strength, E = 3.81 X 10^9 NC^-1

4. F = qE

F = (2.0 X 10^-3)(100) = 0.2 N

5. The electric field from the long wire to a certain distance is inversely proportional to the distance

So, E1 / E2 = d2 / d1

(1900) / E2 = 38 / 22

Required field strength, E2 = 1100 NC^-1

6. Electric force balances the weight of the mass drop

qE = mg

(10)(1.6 X 10^-19)(20 X 10^6) = m(10)

mass, m = 3.2 X 10^-12 kg

7. Assume Newtonian mechanics is still valid.

One-tenth of speed of light = 3.0 X 10^7 ms^-1

By equation of motion, v^2 = u^2 + 2as

(3.0 X 10^7)^2 = 0 + 2a(0.05)

Acceleration, a = 9 X 10^15 ms^-2

By Newton's 2nd law of motion,

F = ma

F = (9.11 X 10^-31)(9 X 10^15)

Electric force, F = 8.2 X 10^-15 N

By F = qE

E = F / q = (8.2 X 10^-15) / (1.6 X 10^-19) = 5.12 X 10^4 NC^-1

參考: Physics king

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