Two conducting plates connected to the terminals of a battery (of voltage V0) ar

Question

Two conducting plates connected to the terminals of a battery (of voltage V0) are a distance D apart. The inner surface of the left plate has a charge density of -?, the inner surface of the right plate has a charge density of +?. An uncharged metal slab of width w is placed between the plates.

1. Describe the electric field inside the metal slab. What charge distribuion would account for this field?

2. Write an equation in terms of V0 for the amount of work that must be done on a particle of charge qo to move it from the plate on the left (-?) to the plate on the right (+?).

3. Does intruducing the metal slab change the electric field in the region outside the conducting plates? Between the plates (but outside the slab)? Explain.

4. Does introducing the metal slab change the distribution on the conducting plates?

Explanation / Answer

a) E ? =0 as inside the metal electric field is zero... metal slab face which is near to +vely charged plate will have ?? sufrace chage distribution and the face which is near to -vely charged plate will have +? surface charge density . b) W=?q V. c)no nothing will be changed ....new charge distribution of metal slab give zero electric field outside the slab ..so their is no change due to introduction of slab. d)no ...their is no change in charge distribution on metal plate ..

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