Draw the free body diagram for the charge C showing the superposition of forces FC= FA + FB. Calculate the net force (magnitude and direction) on the charge labeled C at the far right.
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Draw the free body diagram for the charge C showing the superposition of forces FC= FA + FB.
Calculate the net force (magnitude and direction) on the charge labeled C at the far right.
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- For the diagram shown below, find the magnitude and direction of the net force on the charge on the lower right corner of the square (q = 6 µC, a=1 cm) Free-body diagram with vectors force to scale Magnitude and directionA Three point charges each of magnitude 11.0 µC are located at the corners of an equilateral triangle of side 15.0 cm as shown. (a) Sketch and label the forces exerted on each charge. {b} Calculate the magnitude and direction of the net force exerted on Q3. (c) Why does a glass rod become positively charged when rubbed with silk?There could be charge on C, but there is no 1. net charge on C. 2. There cannot be charge on C. There could be charge on C and there is net 3. charge on C. C A metal block containing a cavity and car- rying a net charge –Q is located near a pos- itive charge +q_as shown in the figure. In static equilibrium, can there be any charges on the inner surface C of the cavity?
- Three charges are arranged as shown in the figure below. Find the magnitude and direction of the electrostatic force on the charge q = 4.76 nC at the origin. (Let r12 = 0.240 m.)Three point charges lie along the axes in the x y-coordinate plane.Positive charge q is at the origin.A charge of 6.00 nC is at (r1 2, 0), where r1 2 > 0.A charge of −3.00 nC is at (0, −0.100 m).The diagram at the right shows three charges positioned to form an equilateral triangle. Each side has a length of 46 cm and each charge has a positive charge of 7.8 nC. Determine the magnitude and direction of the net electric force exerted upon the charge at point P at the top of the triangle.Why is C the answer?I need help figuring out this problem.
- 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.This applet helps you practice the drawing of a free body diagram for a charge experiencing electric forces fi other charges. Three charges 9₁, 92 and q3 are placed along a staright line as shown below. The charge q2 is placed halfway between q1 and q3 separated by a distance 2d. +4uC +2μC +6μС 91 = +4μC F1-2 F1-2 is the force of q₁ on 92. F3+2 is the force of q3 on q2- In the applet below, select the force vectors on q₂ due to q₁ and 93. Place the tails of the force vectors on 92 For each force, there are four choices, two specifying the directions of the forces and two specifying the magnitudes of the forces. If F1 2 > F3+2, then pick the longer vector for F₁2 and the shorter vector for F3-2, and vice-versa. How will you determine which force is greater? F1-2 d F₁ 2 d F1-2 Select the appropriate directions and relative magnitudes of F1-2 and F3-2 and place their tails on the test charge, 92. 92 = +2 μC X F3-2 93 = +6 μC F3, 12 F3-2 F3-2help