Imagine a night angle triangle with two equal sides of 10 cm Two equal charges of +10uC are placed in the vertices corresponding to the 45° angles A negative, 20µC charge is placed at the 90° vertex Find the electnic field in the middle of the hypotenuse. You need to provide the magnitude of the field and the direction. For the direction use the angle between the electic field and the hypotenuse (the one less than 90) Part A - What is the magnitude of the field at the point in the midle of the hypothenuse? Value Units Submit Request Answer Part B. What is the angle between the total field and the hypothenuse? Give the angle which is less than 90 degrees.

Imagine a night angle triangle with two equal sides of 10 cm Two equal charges of +10uC are placed in the vertices corresponding to the 45° angles A negative, 20µC charge is placed at the 90° vertex Find the electnic field in the middle of the hypotenuse. You need to provide the magnitude of the field and the direction. For the direction use the angle between the electic field and the hypotenuse (the one less than 90) Part A - What is the magnitude of the field at the point in the midle of the hypothenuse? Value Units Submit Request Answer Part B. What is the angle between the total field and the hypothenuse? Give the angle which is less than 90 degrees.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

-q -q +q In which direction will the resultant or net force vector from the two equal negative charges move the negative charge shown in the figure? Note, the angle between the two pairs of positive and negative charges is the same with respected to the dotted horizontal line. Hint: Think about the components of the Coulomb force due to each negative charge. The net force due to both negative charges will move the positive charge to the east. The net force due to both negative charges will move the positive charge to the west. The net force due to both negative charges is 0, so the positive charge remains at rest. The net force due to both negative charges will move the positive charge to the northeast.
A charge q = 3.0 pC is placed at the point P shown below. What is the force on a (in N)? (Assume the positive direction is to the right. Indicate the direction with the sign of your answer.) 1.9 m -3.0 uc 3.8 m 1.0 uC 22.6 10 x N
Three equal point charges of varying signs are placed on three of the corners of a square of side d as shown in Figure. Which of the arrows shown represents the direction of the net electric field at the vacant corner of the square? d -q +0 A B D d none of the above A 8 -q D
Three point charges are arranged as shown in the figure below. (a) Find the electric field vector (magnitude and direction) the 6.00 nC and -3.00 nC charges create at the origin. (Ignore the 5.00 C charge for now!). (b) Find the force vector (magnitude and direction) on the 5.00 nC charge when it is placed at the orgin. (Hint: use your answer from part (a) to make this easier.)
Three charges are located at different positions on a 2-dimensional grid. There is a +5 uC charge at the origin, a +2uC charge on the x axis at x=2m, and a -4uC charge on the y axis at the point y=3m. Find the forces on each of the three charges.
Please answer both.
Three-point charges are arranged as shown in the figure below. Find the magnitude and direction of the electric force on the particle q = 5.20 nC at the origin. (Let r12 = 0.260 m.) find magnitude and direction counterclockwise from the +x axis
Three point charges are aligned along the x axis as shown in Figure. Find the vector force on the 5.00-nC charge. Write your answer in terms of unit vectors. n=10^(-9) k=8,99*10^9 N/(m^2*C^2)
Number 1. Three charges are at the corners of an isosceles triangle as shown in the figure below. The +5.00 – µC charges form a dipole. (a) Find the force (magnitude and direction) the -10.00 – µc charge exerts on the dipole. (b) For an axis perpendicular to the line connecting the +5.00µC charges at the midpoint of this line, find the torque (magnitude and direction) exerted on the dipole by the -10.00µc charge. +5.00 µC 2.00 cm 3.00 cm -10.00 μC 2.00 cm 5.00 μ C
In this problem we explore dipoles as sources of fields and their interactions with other charges An induced dipole is a dipole (separated, opposite charges) whose charge separation and dipole moment are caused by the presence of an external electric field, often due to some other source charge. Often, the induced dipole moment is proportional to that electric field at the location of the dipole. For example, when a charged piece of tape is near your finger, the charges inside the neutral atoms in your finger move in response to the tape's field at the location of the finger, and each atom in the finger becomes a dipole. The closer the tape is to the finger, the stronger the tape's field, the more the charges move and the larger the dipole moments of the atoms in the finger. Mathematically, we can write this as p = aE, where p is the dipole moment, E is the strength of the electric field at the location of the dipole, and a is a constant that depends on the type of atom and gives the…
Anil
Three negative point charges lie along a line as shown in the figure below. Find the magnitude and direction of the electric field this combination of charges produces at point P, which lies 6.00 cm from the -2.00 µC charge measured perpendicular to the line connecting the three charges. -500 μC 8.00 cm 6.00 cm - 2.00 μC 8.00 cm - 5.00 µC Find the magnitude and direction of the electric field this combination of charges produces at point P, which lies 6.00 cm from the -2.00 µC charge measured perpendicular to the line connecting the three charges. * 1.04 x10' N/C, toward the -2.00 µC charge 1.04 x10' N/C, toward the upper -5.00 µC charge 1.04 x10' N/C, toward the bottom -5.00 µC charge