a). what is the electric field created by two parallel plates which are placed: 1). 3cm apart attached to 150v battery, 2). 5cm apart attached to 350v battery, 3). 1cm apart attached to 450v battery? b) in each case, what is the electrical potential AND the electrical potential energy of a +1.5 micro coulomb charge placed in the middle of the two plates? c). in each case, what is the electrical potential AND electrical potential energy of a +1.5 micro coulomb charge placed 1/4 of the distance from the positive plate?
ANY HELP......Some tough stuff
ANY HELP......Some tough stuff
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a) The concept is pretty simple, but there are certain assumptions that are given. The first assumption is that the plates are infinitely wide, which is a common and accurate assumption except near the ends of the plates. Another assumption is that the metals are equipotential, or that everywhere on a piece of metal without anything separating pieces, it has the same voltage. This is also an accurate assumption unless the voltage wavelength is short enough to appear on the metal (you don't have to think about this, but if you study RF signals then that becomes important).
Anyway, start with the definition of the relationship of voltage and electric fields -- that is, E = -dV/dx, and V = -integral E dx (in three dimensions then differentiation becomes gradient, and integral becomes vector contour integral, but the above approximations make the y and z directions go away). Given the above assumptions, you can assume that the electric field E is constant (this is the large plate assumption and the equipotential metal assumption), which means that E = V/d, where d is the distance. This covers all your information for a).
b), c) Electric potential is E, electric potential energy is q*V. Since you know E is constant and V is therefore linearly scaled from one plate to the next, you already know the answers.
Sorry if there's too much information. In a physics class at your level, those assumptions are made for every problem you ever do regarding capacitance. In my electromagnetics classes, we drop a lot of those assumptions.
Anyway, start with the definition of the relationship of voltage and electric fields -- that is, E = -dV/dx, and V = -integral E dx (in three dimensions then differentiation becomes gradient, and integral becomes vector contour integral, but the above approximations make the y and z directions go away). Given the above assumptions, you can assume that the electric field E is constant (this is the large plate assumption and the equipotential metal assumption), which means that E = V/d, where d is the distance. This covers all your information for a).
b), c) Electric potential is E, electric potential energy is q*V. Since you know E is constant and V is therefore linearly scaled from one plate to the next, you already know the answers.
Sorry if there's too much information. In a physics class at your level, those assumptions are made for every problem you ever do regarding capacitance. In my electromagnetics classes, we drop a lot of those assumptions.