It is argued that all the methods for solving Laplace’s equation carry over to linear dielectrics, with a few small tweaks. Briefly describe what modifications are needed in the general boundary conditions for electric field and electric potential when dielectrics are present (do not merely rewrite the equations, but rather describe what quantities we need to account for being different in our calculations).

It is argued that all the methods for solving Laplace’s equation carry over to linear dielectrics, with a few small tweaks. Briefly describe what modifications are needed in the general boundary conditions for electric field and electric potential when dielectrics are present (do not merely rewrite the equations, but rather describe what quantities we need to account for being different in our calculations).

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Imagine that two parallel plates are charged to achieve a voltage difference Vo between the plates. They have area A and spacing d. The plates are then electrically isolated so that no charge can be added or taken away. A good conductor of thickness d/2 is placed in between the two plates, as shown. What then is the capacitance of this new configuration? Is it different from the original capacitance?
I keep getting different answers with each calculation and unsure of where I am going wrong. If possible, please show steps so I may understand where my calculations were incorrect. (a). What's the capacitance of your capacitor? (the value and unit(s)). (b). If you disconnect the battery and separate the plates to a distance of 3.50 cm without discharging them, what will be the potential difference between them? (the value and unit(s)).
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An ideal air-filled parallel-plate capacitor holds a fixed amount of equal but opposite charge on its plates. All the geometric parameters of the capacitor (each plate diameter and plate separation) are now HALVED. What happens to the each of these quantities: (circle the correct option, and write the change factor, if applicable. Show work.) a) capacitance, C: stays the same changes by a factor of _____ b) energy density, u: stays the same changes by a factor of _____ c)voltage difference across plates, V: stays the same changes by a factor of _____ d) electric field between the plates, E stays the same changes by a factor of _____
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Consider a thin, uniformly charged rod of length L with total charge Q and test points A, a distance a from the center of the rod and B a distance b from the rod. В Find the potential difference between A and B first by integrating the point source potential to find V and V and subtracting, and then by integrating the field. Compare the results in the limit of L>>(a and b). To test the far field limit, compare the appropriate result to the case where L is much less than both a and b. You may need to do this one numerically.
How do you do this one? Specifically, for b), the answer gives Vb > Va, why is that? In general how do we know which side has higher potentials? (notice: this is not a graded question, it's a practice problem that comes with an answer which I don't quite understand, you can see from how I asked the question that the answer is provided...)