I need just answer no need solution. The figure shows two equal positive charges91 = 92 = Q, located at (x, y) = (±a,0) and athird charge q3 = -2Q at (x, y) = (0,-2a).%3D%3DFind the total electric field vector E, at thea)point P which is assumed to place at(x, y) = (0, a).%3Db) If a fourth charge q4 = -Q is put at P, findthe total force acting on it.

Answered: gure shows two equal positive charges 2…

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)...

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I need just answer no need solution.

The figure shows two equal positive charges
91 = 92 = Q, located at (x, y) = (±a,0) and a
third charge q3 = -2Q at (x, y) = (0,-2a).
%3D
%3D
Find the total electric field vector E, at the
a)
point P which is assumed to place at
(x, y) = (0, a).
%3D
b) If a fourth charge q4 = -Q is put at P, find
the total force acting on it.

Expert Solution

Step 1

Given:

$q_{1} = q_{2} = Q$

$q_{3} = - 2 Q$

Step 2

(a) Electric field at $P (0, a)$ because of $q_{1} (- a, 0) a n d q_{2} (a, 0)$ is,

$\vec{E_{P_{1}}} = \sqrt{{(E_{a})}^{2} + {(E_{- a})}^{2} + 2 E_{a} E_{- a} \cos 90 °} \vec{E} = \sqrt{{(k \frac{Q}{r^{2}})}^{2} + {(k \frac{Q}{r^{2}})}^{2}} \vec{E_{P_{1}}} = \sqrt{{(k \frac{Q}{{(\sqrt{2} a)}^{2}})}^{2} + {(k \frac{Q}{{(\sqrt{2} a)}^{2}})}^{2}} = \sqrt{2 {(k \frac{Q}{{(\sqrt{2} a)}^{2}})}^{2}} \vec{E_{P_{1}}} = \sqrt{2} (k \frac{Q}{{(\sqrt{2} a)}^{2}}) = \sqrt{2} k \frac{Q}{2 a^{2}} = \frac{k Q}{\sqrt{2} a^{2}}$

Electric field at P(0,a) due to $q_{3} (0, - 2 a)$ is:

$\vec{E_{P_{2}}} = k \frac{- 2 Q}{r^{2}} = - \frac{k (2 Q)}{{(3 a)}^{2}} = - \frac{k (2 Q)}{9 a^{2}} = - \frac{2}{9} \frac{k Q}{a^{2}}$

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