Electrostatic field due to all three charges can be calculated using Columb's formula.
Then we find the algebraic sum of the field components is x and y-direction and find the resultant field.
The direction of the resultant field is also calculated.
6.00 uC, 92 = 6.00 4C, and q3 = -2.00 uC.) Three charges are at the...
Three charges are at the corners of an equilateral triangle, as shown in the figure below. Calculate the electric field at a point midway between the two charges on the x-axis. (Let 91 = 9.00 uc, 92 = 6.00 uc, and 93 = -3.00 uC.) 0.500 m 60.00 magnitude 1.85e12 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. N/C 18.5 ✓ below the +x-axis direction
Three charges are at the corners of an equilateral triangle, as shown in the figure below. Calculate the electric field at a point midway between the two charges on the x-axis. (Let q1-6.50pC q2.5.50pC, and q,--4.00,rc.) 0.500 m 60.0 magnitude N/C direction below the +x-axis
Three charges are at the corners of an equilateral triangle, as shown in the figure below. Calculate the electric field at a point midway between the two charges on the x-axis. (Let magnitude N/C direction below the +x-axis 0.500 m 60.0
Three charges are at the corners of an equilateral triangle, as shown in the figure below. Calculate the electric field at a point midway between the two charges on the x-axis. (Let q1 = 6.50 pC q2 : 2.50 μC, and q3 =-7.50 μC.) magnitude 1444991レN/C 4.76 × direction Your response differs from the correct answer by more than 10%. Double check your calculations. below the +x- axis 0.500 m 60.0
1. Charges 91 = +1.00 uC, 42 = -2.00 uC and 43 = +3.00 uC are placed on three corners of a square of side a = 0.500 m as shown in the diagram. (a) On the diagram, draw the electric field vectors at point P (the fourth corner) of the three charges. Label the electric field of qı as Eį, etc. (b) Calculate the net electric field at P and write it in vector notation. (c) Calculate the magnitude...
Three charges are at the corners of an equilateral triangle, as shown in the figure below. Calculate the electric field at a point midway between the two charges on the x-axis. (Let q1: 7.00 μC, q2-7.00 μC, and q,--4.50 μC.) magnitude 169.196 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. N/C 12.1 Your response is within 10% of the correct value. This may be due to roundoff error, or...
Three charges are at the corners of an equilateral triangle, as shown in the figure below. Calculate the electric field at a point midway between the two charges on the x-axis. (Let q1 = 6.50 μC,q2 = 2.00 μC, andq3 = −2.50 μC.)magnitude N/Cdirection ° below the +x-axis
Three charges, q1 = +2.61nC, q2 =-1.41nC, and q3 = +9.41nC, are at the corners of an equilateral triangle, each of whose sides are of length, L, as shown in the figure below. 92 The angle a is 60.0° and L- 0.429 m. We are interested in the unmarked point midway between the charges q1 and q2 on the x axis. For starters, calculate the magnitude and direction of the electric field due only to charge q1 at this point....
Three point charges, 91 = 4.00 PC, 92 = 2.00 uc, and Q = -5.00 PC are located on the vertices of a right triangle as shown in the figure below. The distances du = 4.0 cm and d2 = 2.0 cm. (a) Calculate the magnitude of the electric field at position of the charge Q due to the charge qı? N/C (b) Determine the direction of the electric field. toward the bottom of the screen toward the top of...
0/2 points 1 Previous Answers SerCP9 15.P028 M Three charges are at the corners of an equilateral triangle, as shown in the figure below Caloulate the electric field at a point midway between the two charges on the x-axis. (Let 1-8.00 JC, 2-5.00 HC, and 93--7,00 uc 19019 magnitude Your response differs significantly from the correct answer, Rework your solution from the beginning and check each step carefully. N/c direction 43 eck your alculationse below the +x-axis Your response dimers...