Two small charged metal spheres A and B are situated in a vacuum. The distance between the centres of the spheres is 12.0 cm, as shown in Fig. 4.1. enter image description here

The charge on each sphere may be assumed to be a point charge at the centre of the sphere. Point P is a movable point that lies on the line joining the centres of the spheres and is distance x from the centre of sphere A. The variation with distance x of the electric field strength E at point P is shown in Fig. 4.2.

enter image description here

1. How to know that the spheres are conductors ??
2. How to know that the charges are both positive or both negative ??

For part1: electric field strength change thus it conducts.
For part2: as electric field strength change from positive to negative, they are both positive.
It is correct ??


closed as off-topic by John Rennie, JamalS, Kyle Kanos, DavePhD, Jim Oct 7 '14 at 13:11

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  • $\begingroup$ Hello. You must say what you have tried $\endgroup$ – soumyadeep Oct 7 '14 at 6:38
  • $\begingroup$ Just added, is it correct $\endgroup$ – Arodi007 Oct 7 '14 at 6:42
  • $\begingroup$ Are you trying to reason out with physical sense?Can be painful.I suggest you write the equations for each each case and draw the graphs separately.The one that matches gives you the right combination. $\endgroup$ – soumyadeep Oct 7 '14 at 6:45
  • $\begingroup$ I actually need to use physical sense in my exam question... $\endgroup$ – Arodi007 Oct 7 '14 at 6:48
  • $\begingroup$ As you are new to this site-Check my work type of questions are discouraged here.You question is very close to that.So you must not directly ask in such a way.You can modify your question in such a way that it asks about a physics concept and answers your work indirectly. $\endgroup$ – soumyadeep Oct 7 '14 at 7:14

1.You are or more precisely the graph is considering the distance between the spheres ,where $ \vec E$ for both conducting and non conducting sphere is same.So the graph is same irrespective of whether its conducting or not.So the first answer is wrong.

2.As from the graph,at the midpoint the field is 0,it is clear that both have the same charge of same magnitude and sign.Now,on the left of the midpoint,the left sphere will get priority as the $r$ in the denominator of $\frac {q}{4\pi \epsilon r^2 }$ will reduce and its value will be greater.Now we see on the left of midpoint $\vec E$ is positive i.e towards the right.As the left sphere gets more priority, it must repel the test charge on the right-So the left sphere must be positively charged.Similarly the case for the right one.


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