Below is a positively charged conducting sphere of radius R, bèlow the sphere is shown the variation of a physical property, f(r), as a function of radial distance r from the center of the charged conducting sphere. Identify the physical quantity, f(r), that varies as shown below. (Give one line answer in the blank provided, Explain your answer in the procedure sheet) +, R+ + f (r)
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Below is a positively charged conducting sphere of radius R, below the sphere is shown the variation of a physical property, f(r), as a function of radial distance r from the center of the charged conducting sphere. Identify the physical quantity, f(r), that varies as shown below.
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- Please do all partsA solid disk of radius R = 11 cm lies in the y-z plane with the center at the origin. The disk carries a uniformly distributed total charge Q = 45 μC. A point P is located on the positive half of the x-axis a distance 24 cm from the origin. Refer to the diagram, where the y- and z-axes lie in the plane of the screen and the x-axis points out of the screen. a. Enter an expression for the surface charge density σ in terms of the total charge and radius of the disk. σ = b. Consider a thin ring of the disk of width dr located a distance r from the center. Enter an equation for the infinitesimal charge in this thin ring in terms of Q, R, r, and dr. dQ = c. Calculate the electric potential at P, in kilovolts. V = d. Calculate the magnitude of the electric field at the point P in units of meganewtons per coulomb. E =The figure below shows a small, charged bead, with a charge of q = +41.0 nC, that moves a distance of d = 0.174 m from point A to point B in the presence of a uniform electric field E of magnitude 255 N/C, pointing right. A positive point charge q is initially at point A, then moves a distance d to the right to point B. Electric field vector E points to the right. (a) What is the magnitude (in N) and direction of the electric force on the bead? magnitude Ndirection (b) What is the work (in J) done on the bead by the electric force as it moves from A to B? J (c) What is the change of the electric potential energy (in J) as the bead moves from A to B? (The system consists of the bead and all its surroundings.) PEB − PEA = J (d) What is the potential difference (in V) between A and B? VB − VA = V
- A negative point charge Q1, is located at the origin. A rod of length L is located along the x axis with the near side a distance d from the origin. A positive charge Q2, is uniformly spread over the length of the rod. After integrating the force from each slice over the length of the rod, the magnitude of the electric force on the charge at the origin can be represented as the following: F = (k |Q1| |Q2|) / (d (d + L)) Let L = 2.22m, d = 0.42m, Q1 = -6.29µC, and |Q2| = 11.1µC. Calculate the magnitude if the force in newtons that the rod exerts on the point charge at the origin.Two spherical, hollow conductors are concentrically nested as shown in the cross-sectional diagram below and electrically isolated from each other. A net charge of -3 nC is divided between the conductors, with a total of -12 nC on the inner conductor and +9 nC on the outer one. The charges are, of course, free to move between the surfaces within each shell but cannot move from one shell to the other. The inner conductor has an inner radius of a=2 cm, outer radius of b=3 cm. The outer conductor has an inner radius of c=6 cm, an outer radius of d=8 cm. a (a) In equilibrium, all of the excess charge is found on the surfaces. Why is there no charge between a and b, nor between c and d? (b) Determine the net charge on each surface (ie how much charge is there at r = a, at r = b, at r = c and at r= = d). (c) Is the charge on the outer surface equal to the net charge on the outer container, the net charge of the system, or some other amount? Explain briefly. (d) Find the surface charge…The figure below shows a charged particle, with a charge of q = +38.0 nC, that moves a distance of d = 0.185 m from point A to point B in the presence of a uniform electric field E of magnitude 245 N/C, pointing right. A positive point charge q is initially at point A, then moves a distance d to the right to point B. Electric field vector E points to the right. (a) What is the magnitude (in N) and direction of the electric force on the particle? magnitude Ndirection ---Select--- toward the right toward the left The magnitude is zero. (b) What is the work (in J) done on the particle by the electric force as it moves from A to B? J (c) What is the change of the electric potential energy (in J) as the particle moves from A to B? (The system consists of the particle and all its surroundings.) PEB − PEA = J (d) What is the potential difference (in V) between A and B? VB − VA = V
- È= [dE = [k9 .. Start with... f... dxA 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 0/99The figure below shows a small, charged bead, with a charge of q = +42.0 nC, that moves a distance of d = 0.189 m from point A to point B in the presence of a uniform electric field E of magnitude 270 N/C, pointing right. A positive point charge q is initially at point A, then moves a distance d to the right to point B. Electric field vector E points to the right. (a) What is the magnitude (in N) and direction of the electric force on the bead? magnitude Ndirection (b) What is the work (in J) done on the bead by the electric force as it moves from A to B? J (c) What is the change of the electric potential energy (in J) as the bead moves from A to B? (The system consists of the bead and all its surroundings.) PEB − PEA = J (d) What is the potential difference (in V) between A and B? VB − VA = V