(d) With minimal reanalysis (just note where you need to change the work you did above and how that changes the result) find the field at the test point if we replace the charge on half the rod with the opposite charge, so we have +Q/2 on a length L/2 and -Q/2 on the remaining length L/2. R (e) Show that the results with this new distribution, in the L << R limit is consistent with a dipole that has dipole moment p=(Q/2) (L/2) = QL/4 and is oriented tangent to the arc (perpendicular to R). You will want to use the correct small angle approximation for cosine (see the exercises!)
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- A uniformly charged rod of length L is bent into an arc of radius R. Total charge on the rod is Q. This is the arc problem we considered in class. R (a) Find the angle in radians, 0, subtended by the arc and the linear charge density in terms of Q, L and R. See the exercises and the hints and guide at the end of the homework for a reminder of the relationship between angle and arc length. (b) Find the field at a test point placed at the center of the circle that the arc is part of. Again, follow the process, and remember that the sample charge dq occupies a small length of the rod, and that length is now a small part of an arc. (c) Check that your results for the case L << R give you the field of a point source with charge Q, a distance R away. You will want to use the small angle approximation from the exercises: sin for 0 << 1 (d) With minimal reanalysis (just note where you need to change the work you did above and how that changes the result) find the field at the test point if we…A uniformly charged rod of length L is bent into an arc of radius R. Total charge on the rod is Q. This is the arc problem we considered in class. ( R P (a) Find the angle in radians, 0, subtended by the arc and the linear charge density λ in terms of Q, L and R. See the exercises and the hints and guide at the end of the homework for a reminder of the relationship between angle and arc length. (b) Find the field at a test point placed at the center of the circle that the arc is part of. Again, follow the process, and remember that the sample charge dq occupies a small length of the rod, and that length is now a small part of an arc. (c) Check that your results for the case L << R give you the field of a point source with charge Q, a distance R. away. You will want to use the small angle approximation from the exercises: sin for 0 << 1 (d) With minimal reanalysis (just note where you need to change the work you did above and how that changes the result) find the field at the test point…Problem 1: A spherical conductor is known to have a radius and a total charge of 10 cm and 20uC. If points A and B are 15 cm and 5 cm from the center of the conductor, respectively. If a test charge, q = 25mC, is to be moved from A to B, determine the following: What isThe work done in moving the test charge;
- Consider the following charge configurations below. Here q, = 5 µC, q2 = 5 µC, and q3 = 10 µC. The distance is a = 1 mm. Calculate the force on 93. You should calculate the force component along x and y directions). Present the results in terms of a force vector. aProblem 1: A spherical conductor is known to have a radius and a total charge of 10 cm and 20uC. If points Aand B are 15 cm and 5 cm from the center of the conductor, respectively. If a test charge, q = 25mC, is to bemoved from A to B, determine the following: c. The work done in moving the test chargeA charge q is placed in the cavity of a conductor as shown below. Will a charge outside the conductor experience an electric field due to the presence of q? The conductor in the preceding figure has an excess charge of -5.0 uC. If a 2.0 uC point charge is placed in the cavity, what is the net charge on the surface of the cavity and on the outer surface of the conductor?
- A uniformly charged rod of length L and total charge Q lies along the x axis as shown in in the figure below. (Use the following as necessary: Q, L, d, and ke.) (a) Find the components of the electric field at the point P on the y axis a distance d from the origin. (b) What are the approximate values of the field components when d >> L?Find the ratio of q/Q for the E-field to be zero at adistance of z = 3.20R for the charge distributionand geometry of problem #30 of the text. a isthe charge on the LARGER ring. Q is the chargeon the SMALLER ring. Answer in 5 Significant Figures!!Assume a uniformly charged ring of radius R and charge Q produces an electric field E at a point Pon its axis, at distance x away from the center of the ring as in Figure a. Now the same charge Q is spread uniformly over the circular area the ring encloses, forming a flat disk of charge with the same radius as in Figure a. How does the field Eick produced by the disk at P compare with the field produced by the ring at the same point? O O Ek Ering O impossible to determine
- A positive charge q is fixed at point (−4, 3) and a negative charge -q is fixed at point (-4,0). Determine the net electric force Fnet acting on a negative test charge -Q at the origin (0, 0) in terms of the given quantities and physical constants, including the permittivity of free space Eo. Express the force using ij unit vector notation. Enter precise fractions rather than entering their approximate numerical values. F net = = 4- 3- 2- 1- -1- -2- -3- y 2 4vt . A line of positive charge is formed into a semicircle of radius R as shown in the fig- ure to the right. The charge per unit length along the semicircle is given by A, and is constant. The total charge on the semicir- cle is Q. R (a) Determine the constant, A, in terms of the Coulomb constant k, total charge Q, and radius R. (b) What is the electric ficld, E (magnitude and direction), at the origin (the center of curvature)? Note: Recall that the arclength subtended by an angle 0 in radians along a circle of radius R is s = R0. Furthermore, you might find the following integral useful: "T/2 cos 0 do = 2. T/2Gauss’s law also helps us locate charge on conductors. There are several distinct pieces we need forthis, listed below. Explain briefly why is each of them true.(a) The electric field inside a conductor (in the conducting material) is zero if the conductor is instatic equilibrium (charges are not moving or at least not accelerating.)(b) The total charge on an isolated conductor (or, really, any object) is constant.(c) The net charge on the wall of a cavity in a conductor is equal to the opposite of the total chargethat is floating inside the cavity.(d) Any excess charge on a conductor is all at the surfaces - including both the outer surface of theconductor and any inner surfaces of cavities - but never in the material.(e) The charge on the outer surface of a conductor is equal to the total charge on the conductorand inside any cavities in it.