▼Part AA charge 5.01 nC is placed at the origin of an xy-coordinatesystem, and a charge -2.00 nC is placed on the positive x-axis atx = 3.96 cm . A third particle, of charge 6.02 nC is now placed atthe point x = 3.96 cm , y = 2.96 cm .Find the x-component of the total force exerted on the third charge by the other two.Express your answer in newtons.ΗV ΑΣφ?Fa =N%3DSubmitRequest AnswerPart BFind the y-component of the total force exerted on the third charge by the other two.Express your answer in newtons.?Fy =N

Answered: Part A A charge 5.01 nC is placed at…

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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Question

▼
Part A
A charge 5.01 nC is placed at the origin of an xy-coordinate
system, and a charge -2.00 nC is placed on the positive x-axis at
x = 3.96 cm . A third particle, of charge 6.02 nC is now placed at
the point x = 3.96 cm , y = 2.96 cm .
Find the x-component of the total force exerted on the third charge by the other two.
Express your answer in newtons.
ΗV ΑΣφ
?
Fa =
N
%3D
Submit
Request Answer
Part B
Find the y-component of the total force exerted on the third charge by the other two.
Express your answer in newtons.
?
Fy =
N

Expert Solution

Step 1

Given:

$Charge, q_{1} = 5.01 nC at origin Charge, q_{2} = - 2.0 nC at x = 3.96 cm Charge, q_{3} = 6.02 nC at x = 3.96 cm, y = 2.96 cm$

Solution:

The electrostatic force exerted on charge q₃ due to charges q₁ and q₂ will be given as:

$\vec{F_{3}} = \vec{F_{31}} + \vec{F_{32}}$

$where, positon vectors \vec{r_{31}} and \overset{⇀}{r_{32}} are as follows : \vec{r_{31}} = 3.96 \hat{i} + 2.96 \hat{j} and \overset{⇀}{r_{32}} = 2.96 \hat{j}$

( i ) For force exerted by charge 1 on charge 3 is given as:

$\begin{array}{rcl} \vec{F_{31}} & = & k \frac{q_{1} q_{3}}{{|\vec{r_{31}}|}^{3}} \cdot \vec{r_{31}} \\ Substituting values in above eqaution,we get: \\ \vec{F_{31}} & = & (9 \times 10^{9} {Nm}^{2} / C^{2}) \frac{(5.01 \times 10^{- 9} C) \times (6 . 0210^{- 9} C)}{{(4.944 \times 10^{- 2})}^{3}} \cdot (3.96 \hat{i} + 2.96 \hat{j}) \\ = & [(8.32 \times 10^{- 3}) \hat{i} + (6.78 \times 10^{- 3}) \hat{j}] N \end{array}$

( ii ) Force force exerted by charge 2 on charge 3 is given as:

$\begin{array}{rcl} \vec{F_{32}} & = & k \frac{q_{2} q_{3}}{{|\vec{r_{32}}|}^{3}} \cdot \vec{r_{32}} \\ Substituting values in above eqaution,we get: \\ \vec{F_{32}} & = & (9 \times 10^{9} {Nm}^{2} / C^{2}) \frac{(2.00 \times 10^{- 9} C) \times (6.02 \times 10^{- 9} C)}{{(2.96 \times 10^{- 2})}^{3}} \cdot (- 1) \hat{j} = [1.237 \times 10^{- 4} \hat{j}] N \end{array}$

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