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Positive charge Q is uniformly distributed around a semicircle of radius
a as shown in Fig.. Find the magnitude and direction of the resulting
electric field at point P, the center of curvature of the semicircle.
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- In Figure (a) below, a particle of charge +Q produces an electric field of magnitude Epart at point P, at distance R from the particle. In Figure (b), that same amount of charge is spread uniformly along a circular arc that has radius R and subtends an angle 8. The charge on the arc produces an electric field of magnitude Earc at its center of curvature P. For what value of 0 (in º) does Earc = 0.75Epart? (Hint: You will probably resort to a graphical solution.) +Q |▬▬▬R—-|| Number i P (a) AR +Q}/0/2/ (b) Units ° (degree:Please Asapanswer the following
- A cube has positive charge +Q in all corners except for one, which has a negative point charge - Q. Let the distance from any corner to the center of the cube be r. What is the magnitude and direction of the of electric field at the center of the cube (point P)?A thin, circular disk of radius R = 45 cm is oriented in the yz-plane with its center as the origin. The disk carries a total charge Q = 6.0 μC distributed uniformly over its surface. Calculate the magnitude of the electric field due to the disk at the point x = 12 cm along the x-axis.A negative line of charge is on the negative y axis with its top end at the origin as shown in the figure below. The charge is uniformly distributed along the line. Point P is on the positive y axiS as shown in the figure. y The the length of the line of charge is 3.17 m, the total charge on the line is -787.9 nC and point P is 3.15 m from the origin. Find the magnitude of the electric field at point P due to the line of charge. (in N/C) OA: 3.15×10²| OB: 3.56×102| 0C: 4.02×102| OD: 4.54×102| OE: 5.13×102|| OF: 5.80×102|| OG: 6.56×102|| OH: 7.41×102 Submit Answer Tries 0/99