[3.1] Describe the electric field lines between the plates. (see image below) [3

Question

[3.1] Describe the electric field lines between the plates. (see image below)

[3.2] From this simple discussion in terms of charges and potential difference, can you deduce a relation between C and the surface area of the conducting plates? Explain your reasoning in your answer.

[3.3] Now imagine the parallel plates are placed in an external electric field. What will be different in each pictured case? How does the value of C change?

[3.4] Explain how the dipole aligns itself. What will happen if the initial placement of the stick is exactly anti-parallel to the E-field?

[3.5] Now imagine you have several of such dipoles, and place them regularly between the plates. What will be different now? How does the C change?

Explanation / Answer

3.1 According to gauss law we can consider the plates are in equlibrium. Lets consider the charge density of the plates are sigma and the permittivity of free space or the gap between the plates are epsilon.

Then if there is electric field E in this case we get that E=sigma/epsilon

Therefore for both plates the electric field E=sigma/2*epsilon

3.2 Again we know that according to Gauss law sigma or surface charge density= Q/2A

That means E=Q/2A*2epsilon= Q/4Aepsilon

again capacitance C= Q/V that means C=epsilonA/d

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