E&M electrostatics 4 (urgent)

2011-05-04 2:09 am
1. 'When one coulomb of point charge is placed near a large metal sphere of charge Q, the magnitude of force is kQ/r^2, where r is the distance of the point charge from the centre of the sphere and k is the Coulomb's constant.' Why is this statement false?


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thanks
更新1:

Actually all the answers indicated should be correct (according to the answer key), I marks it just for convenience... For 1, it should be false, or can u tell me your steps why you think it's true provided that r>radius ? For 20, why (1) and (2) are correct?

更新2:

For 21, when the ball touches the container, why is there still +ve charge remaining? Because it is earthed and I think the whole set-up should remain neutral. For the last 1, why a decrease in mass can increase the oscillation of the ball? thanks

回答 (1)

2011-05-04 5:13 am
✔ 最佳答案
1. Are you sure that it is false. It should be true provided that r > radius of sphere2. You got the right answer. P has induced +ve charge and Q has induced -ve charge.

20, Your answer is right. Statement (3) is wrong because the same amount of charge has spreaded over a larger area than before. Hence, this leads to wider separation between electric field lines. Because field line separation represents its intensity, a wider separation indicates a weaker field.

21. Statements (1) and (2) are correct. When the sphere is lowered into the container, -ve charge is induced inside the container and +ve charge on the outside. When the sphere touches the container, what remains is only the +ve induced charge on the outside of the container. Since the electroscope measures the potential given by the +ve induced charges (the effect of the -ve induced charge and the +ve charge on the sphere cancell one another out), hence the leaf divergence doesn't change.

22. Electrons can go through the inside of a solid metal sphere, but can only go along the surface of a hollow mwtal sphere.

18. Let e and m be the charge (magnitude) and mass of an electron. Hence, the charge magnitude of an alpha is 2e, and mass 7000m Let d be the distance from the +ve plate where the two particles meet.
Consider the electron, acceleration a = eE/m, where E is the electric field intensity
use equation of motion: s = ut + (1/2)at^2
(0.03 - d) = (1/2)(eE/m)t^2 ------------------------- (1)

Consider the alpha, a = (2e)E/7000m
hence, d = (1/2)(2eE/7000m)t^2 -------------------- (2)

(2)/(1): d/(0.03-d) = (2/7000)
i.e. 3500d = (0.03 - d)
d = 0.03/3501 m = 8.6 x 10^-6 m
The answer is thus option C


2011-05-03 21:17:15 補充:
sorry, I have missed out the last question.
The frequency increases with the force given by the electric field, and decreases with the mass of the ball (just compare with a spring-mass oscillation system).

2011-05-04 08:31:04 補充:
I cannot write anything more on this post because the "words limit" has been exceeded. Perhaps you could pm me for further clarification.


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