A uniform spherical charge distribution (as shown in the figure below) has a total charge of 63.3 mC and radiusR = 19.0 cm. Find the electric force exerted on an electron placed at r= 0, 9.5 cm, 19.0 cm, and 28.5 cm. (Assume the positive direction is radially outward. Indicate the direction with the sign of your answer.) N r = 0 F = r= 9.5 cm F = N r = 19.0 cm F = N r = 28.5 cm F = R 9+
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- In a rectangular coordinate system a positive point charge q = 6.00 x 10-9 C is placed at the point x = +0.150m, y = 0, and an identical point charge is placed at x = -0.150m, y = 0. Find the x- and y-components, the magnitude, and the direction of the electric field at x = 0.300 m, y = 0. (with precise solution please) A. Ex = 129N/C and Ey = -510N/C with direction of 284 degrees counterclockwise from +x axis B. Ex = 2663.1 N/C and Ey = 0 to the +x direction C. Ex = 0 and Ey = 0 D. none of the aboveI need help with part b, please! Point charges of 30.0 µC and 45.0 µC are placed 0.650 m apart. (a) At what point (in m) along the line between them is the electric field zero? ANSWER: 0.292 m away from the 30.0 µC charge (b) What (in N/C) is the electric field halfway between them? (Enter the magnitude.) ______N/CThe 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
- Consider the symmetrically arranged charges in the figure, in which qa= b = -3.65 µC and qe = qd = +3.65 μC. Determine the direction of the electric field at the location of charge q. down and left up and left down and right E = Oleft down up and right up right Calculate the magnitude of the electric field E at the location of a given that the square is 5.45 cm on a side. N/C 9₂ 9c 9 9aThree point charges are shown below. q₁ = 2.0 µC, q2 = 4.0 µC, and q3 = - 2.0 µC. d₁ = 20 cm and d2 = 10 cm. What is the magnitude of the net electric force on q2, in Newtons? Use K = 9.0 × 10⁹ Nm²/C². Your answer needs to have 2 significant figures, including the negative sign in your answer if needed. Do not include the positive sign if the answer is positive. No unit is needed in your answer, it is already given in the question statement. 91 + d 92 + -d₂ 93The point charges in the figure below are located at the corners of an equilateral triangle 25.0 cm on a side, where q, = +19.0 µC and q. = -5.70 µC. (Assume that the +x-axis is directed to the right.) la (a) Find the electric field at the location of q.. magnitude N/C direction ° (counterclockwise from the +x-axis) (b) What is the force on given that 9a = +1.50 nC? magnitude N direction (counterclockwise from the +x-axis)
- Given the two charged particles shown in the figure below, find the electric field at the origin. (Let q₁ = -30.00 nC and 92 = 9.00 nC. Express your answer in vector form.) N/C E = -4 92 -2 y (cm) 4 2 -2 -4 2 4 91 x (cm)Consider the symmetrically arranged charges in the figure, in which qa 9b = -2.45 µC and q. = qd = +2.45 µC. Determine the direction of the electric field at the location of charge q. O right up and left down and left up and right dn down left down and right Calculate the magnitude of the electric field E at the location of q given that the square is 6.35 cm on a side.Given the arrangement of charged particles in the figure below, find the net electrostatic force on the 9₁ = 5.15-μC charged particle. (Assume q2 = 16.33 µC and 93 = -19.12 μC. Express your answer in vector form.) 7 = X N 91 (-2.00 cm, 0) 93 (1.00 cm, 1.00 cm) 92 (0, -1.00 cm)
- Two positive charges, each 4.18 µC, and a negative charge, -6.36 µC, are fixed at the vertices of an equilateral triangle. Calculate the length of the triangle side if the net electrostatic force on the negative charge is Fnet=29.6 N. Give your answer in SI units. Answer: Choose... +Given the two charged particles shown in the figure below, find the electric field at the origin. (Let q₁ = -24.00 nC and 92 = 8.00 nC. Express your answer in vector form.) E E = N/C -4 -2 y (cm) 92 2 91 4 x (cm)The 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