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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?... 顯示更多 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? 圖片參考：http://farm6.static.flickr.com/5183/5683428392_c582a56cb6.jpg 圖片參考：http://farm6.static.flickr.com/5148/5682861939_c26922ed1b.jpg 圖片參考：http://farm6.static.flickr.com/5063/5683428244_250f0ecac3.jpg If you can't see the pictures clearly, please visit the following websites: http://www.flickr.com/photos/29697027@N07/5682861939/in/photostream/ http://www.flickr.com/photos/29697027@N07/5683428244/in/photostream/ http://www.flickr.com/photos/29697027@N07/5683428392/in/photostream/ thanks 更新: 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
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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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E&M electrostatics 4 (urgent)發問:
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?... 顯示更多 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? 圖片參考：http://farm6.static.flickr.com/5183/5683428392_c582a56cb6.jpg 圖片參考：http://farm6.static.flickr.com/5148/5682861939_c26922ed1b.jpg 圖片參考：http://farm6.static.flickr.com/5063/5683428244_250f0ecac3.jpg If you can't see the pictures clearly, please visit the following websites: http://www.flickr.com/photos/29697027@N07/5682861939/in/photostream/ http://www.flickr.com/photos/29697027@N07/5683428244/in/photostream/ http://www.flickr.com/photos/29697027@N07/5683428392/in/photostream/ thanks 更新: 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
最佳解答:
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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