A thin walled metal spherical shell of radius a has charge q_+0, Concentric with
ID: 1592513 • Letter: A
Question
A thin walled metal spherical shell of radius a has charge q_+0, Concentric with it is a thin-walled metal spherical shell of radius b>a and charge q_b. Find the electric field at points a distance r from the common center, where rr>a r>b In the quark model of fundamental particles, a proton is composed of three quarks: two "up" quarks, each having +2e/3, and "down quark, having charge -e/3. Suppose that the three quarks are equidistant from one another. Take that separate distance to be 1.32 fm and calculate the electric potential energy of the system of only the two up quarks and all three quarks. In the figure, a thin nonconducting rod is bent into a complete circle of radius R. The rod is divided into four equally long part, and each part is given a separate charge that is uniformly across it. The quartet of the rod that starts to the left of the radius labelled R has +Q . Going clockwise from there, the quarter circles have charge -Q, +Q, and -Q, in order. The central perpendicular axis through the ring is a z axis, with the origin at the center of the ring. Draw a picture showing the electric field vectors from each quarter circle at a point on the z-axis Use symmetry arguments to draw the net electric field vector at that point. Draw a picture showing the electric field vectors from each quarter circle at a point on the z-axis if the magnitude of both positive charges is doubled. Use symmetry arguments to draw the net electric field vector at that point in this scenario. Find the magnitude of the electric field at a point z = 12.0 cm above the center of the circle in this second scenario with Q= 1.30 fC and R = 4.00 cm.Explanation / Answer
As inside the spherical shell, electric field is zero and outside will be same as it would be due similar amount of charge placed at the centre of the shell i.e. KQ/r2.Hence,
r<a, this is inside the both the shells, hence E = 0.
b>r>a, this is outside smaller but inside the larger shell, hence E = Kqa/r + 0.
When r>a, it is outside the both the shells, hence E= Kqa/r + Kqb/r where is distance of point from the centre.
As P.E = Kqq'/r for two charge system.
a). Hence , due to two up quarks P.E. = 4/9 Ke2/d = 7.76×10^-14 J.
b). Due to all three P.E will be thrice of the P.E due to two hence P.E. = 2.33×10^-13J
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