Asked by Prince
5. Metal sphere A and B in Fig 1.24A are initially neutral
Coulomb’s law describes the
and are touching. A positive charge rod is brought near A,
electrostatics forces between
but not touching.
small point electric chares q1 and
i) Is A now positive, negative, or neutral? Use both charge
q2 at rest or nearly at rest and
diagrams and words to explain.
separated by distance r.
ii) If you bring your finger near a lightweight, negatively
charged hanging ball, the ball swings over toward your
All Answers 1
Answered by
GPT-5 mini
AI
i) Two metal spheres A and B are initially neutral and touching. When a positively charged rod is brought near A (but not touching), electrons in the connected conductor are attracted toward the rod. Electrons flow from the region near B into A, so:
- The side of the conductor closest to the rod (sphere A) becomes negatively charged (excess electrons).
- The far side (sphere B) is left with a deficit of electrons and becomes positively charged.
So A becomes negatively charged (on the side facing the rod) while B becomes positive. The conductor as a whole remains electrically neutral (total negative = total positive), but the charge is separated (induced polarization). (If you now separate the spheres while the rod is present, A will keep negative charge and B positive.)
Simple diagram (top view):
+ rod near A
rod (+)
→
[A: − − −] — [B: + + +]
ii) A lightweight negatively charged hanging ball is attracted to your neutral finger because of electrostatic induction (polarization). Your finger’s charges rearrange: the free/loosely bound electrons in the finger are repelled away from the surface nearest the ball, leaving a slight excess of positive charge on the finger’s near surface and negative charge farther away. The positive induced charges are much closer to the ball than the induced negative charges, so the attractive force from the nearby positive charges (∝1/r^2) is stronger than the repulsive force from the farther negative charges. The net result is a force pulling the negatively charged ball toward your finger.
- The side of the conductor closest to the rod (sphere A) becomes negatively charged (excess electrons).
- The far side (sphere B) is left with a deficit of electrons and becomes positively charged.
So A becomes negatively charged (on the side facing the rod) while B becomes positive. The conductor as a whole remains electrically neutral (total negative = total positive), but the charge is separated (induced polarization). (If you now separate the spheres while the rod is present, A will keep negative charge and B positive.)
Simple diagram (top view):
+ rod near A
rod (+)
→
[A: − − −] — [B: + + +]
ii) A lightweight negatively charged hanging ball is attracted to your neutral finger because of electrostatic induction (polarization). Your finger’s charges rearrange: the free/loosely bound electrons in the finger are repelled away from the surface nearest the ball, leaving a slight excess of positive charge on the finger’s near surface and negative charge farther away. The positive induced charges are much closer to the ball than the induced negative charges, so the attractive force from the nearby positive charges (∝1/r^2) is stronger than the repulsive force from the farther negative charges. The net result is a force pulling the negatively charged ball toward your finger.
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