Two identical positive charges and a negative charge are fixed at the vertices of an equilateral triangle in vacuum. Write down expression for the net electrostatic force Fnet on the negative charge in terms of forces F between the positive and negative charges. Please use "*" for products (e.g. B*A), "/" for ratios (e.g. B/A) and the usual "+" and "-" signs as appropriate. For trigonometric functions use the usual cos and sin (e.g. cos(40) or sin(50), where 40 and 50 are degrees). Please use the "Display response" button to check you entered the answer you expect. Fnet = (sqrt3)*F
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- The figure below shows three small, charged beads, all lying along the horizontal axis. Bead A, at left, has a 6.90 nC charge. Bead B has a 1.70 nC charge and is 3.00 cm to the right of A. Bead C has a -2.80 nC charge and is 2.00 cm to the right of B. Find the magnitude (in N) and direction of the net electric force on each of the beads. Net Force on A magnitude direction Net Force on magnitude direction 3.00 cm -Select- -Select- B toward the left Net Force on C magnitude direction toward the right -Select- -Select- N v N v 98 9c 2.00 cm-Three charges are arranged as shown in the figure below. Find the magnitude and direction of the electrostatic force on the charge q = 5.28 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). magnitude N direction ° counterclockwise from the +x-axisFour identical charged particles (q = +10.4 µC) are located on the corners of a rectangle as shown in the figure below. The dimensions of the rectangle are L = 61.2 cm and W = 15.7 cm. (a) Calculate the magnitude of the total electric force exerted on the charge at the lower left corner by the other three charges. N(b) Calculate the direction of the total electric force exerted on the charge at the lower left corner by the other three charges. ° (counterclockwise from the +x-axis)
- The figure below shows three small, charged beads at the corners of an equilateral triangle. Bead A has a charge of 1.65 µC; B has a charge of 5.40 µC; and C has a charge of -5.10 µC. Each side of the triangle is 0.500 m long. What are the magnitude and direction of the net electric force on A? (Enter the magnitude in N and the direction in degrees below the +x-axis.) y 0.500 m 60.0° C N magnitude o below the +x-axis 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?A positive charge q is fixed at point (3,4) and a negative charge −q is fixed at point (3,0). Determine the net electric force F→net 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 ε0. Express the force using ij unit vector notation. Enter precise fractions rather than entering their approximate numerical values.
- 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).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 charges +1 μC and +13 μC are placed along the x axis, with the first charge at the origin (x = the second charge at x = +1 m. Find the magnitude and direction of the net force on a -8 nC charge 0) and when placed at the following locations below. Overall Hint a. halfway between the two charges: magnitude of force is direction is Select an answer b. on the axis at x = -0.5 m: magnitude of force is is Select an answer c. at the coordinate (x, y) = (1 m, 0.5 m) (half a meter above the +13 μC charge in a direction perpendicular to the line joining the two fixed charges): Hint for (c) Magnitude of force is degrees below -x axis. mN, and the direction is mN, and the mN, and the direction
- Two identical positive charges and a negative charge are fixed at the vertices of an equilateral triangle in vacuum. Write down expression for the for products net electrostatic force Fnet on the negative charge in terms of forces F between the positive and negative charges. Please use (e.g. B*A), "/" for ratios (e.g. B/A) and the usual "+" and "-" signs as appropriate. For trigonometric functions use the usual cos and sin (e.g. cos(40) or sin(50), where 40 and 50 are degrees). Please use the "Display response" button to check you entered the answer you expect. FnetTwo 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... +Pls asap