Question

2 15 points SePSE10 224.OP013 МУ Notes . Ask Your In the figure below, each charged particle is located at one of the four vertices of a square with side length- . In the figure, A 3,B-4, and c-s, and >o. Bg a) What is the expression for the magnitude of the electric field in the upper right corner of the square (at the location of q)? (Use the following as necessary: q, a, and ke.) Give the direction angle (in degrees counterclockwise from the +x-axis) of the electric field at this location. (counterclockwise from the +x-axis) (o) Determine the expression for the total electric force exerted on the charge q. (Enter the magnitude. Use the following as necessary: q, a, and ke.) Give the direction angle (in degrees counterclockwise from the +x-axis) of the electric force on q (Counterclockwise from the +x-axis) e) What If? How would the answers to parts (a) and (b) change if each of the four charges were negative with the same magnitude? Select all that apply O The electric field would be the same magnitude but opposite O The electric field would be the same magnitude and direction as the O The force would be the same magnitude but opposite direction as O The force would be the same magnitude and direction as the force in direction as the field in part (a) field in part (a) the force in part (b) part (b)

Answer 1

The charge $3q$ located at $\vec{r}_1=a\hat{j}$ , charge$4q$ located at  $\vec{r}_2=0$ (origin) ,charge $5q$ located at $\vec{r}_3=a\hat{i}$ and charge $q$ is located at $\vec{r}=a\hat{i}+a\hat{j}$ .

Electric field at the location of $q$ is

$\vec{E}=k_e\frac{3q(\vec{r}-\vec{r}_1)}{|\vec{r}-\vec{r}_1|^3}+k_e\frac{4q(\vec{r}-\vec{r}_2)}{|\vec{r}-\vec{r}_2|^3}+k_e\frac{5q(\vec{r}-\vec{r}_3)}{|\vec{r}-\vec{r}_3|^3}$

$\vec{E}=k_e\frac{3q(a\hat{i})}{|a\hat{i}|^3}+k_e\frac{4q(a\hat{i}+a\hat{j})}{|a\hat{i}+a\hat{j}|^3}+k_e\frac{5q(a\hat{j})}{|a\hat{j}|^3}$

$\vec{E}=k_e\frac{3q(\hat{i})}{a^2}+k_e\frac{4q(\hat{i}+\hat{j})}{2\sqrt{2}a^2}+k_e\frac{5q(\hat{j})}{a^2}$

$\vec{E}=k_e\frac{q}{a^2}\left ( 3\hat{i}+\sqrt{2}\hat{i}+\sqrt{2}\hat{j}+5\hat{j} \right )$

$\vec{E}=k_e\frac{q}{a^2}\left [ (3+\sqrt{2})\hat{i}+(5+\sqrt{2})\hat{j} \right ]$

$\vec{E}=k_e\frac{q}{a^2}\left [ (4.414)\hat{i}+(6.414)\hat{j} \right ]$

${E}=k_e\frac{q}{a^2}(7.79)$

a)

Expression for magnitude of electric field is ${E}=(7.79)k_e\frac{q}{a^2}$

Direction of electric field is $\theta=\tan^{-1}\left ( \frac{6.414}{4.414} \right )=55.5\degree$ counter-clockwise from positive X-axis.

b)

Electric force exerted on charge q

$\vec{F}=q\vec{E}=k_e\frac{q^2}{a^2}\left [ (4.414)\hat{i}+(6.414)\hat{j} \right ]$

$F=(7.79)k_e\frac{q^2}{a^2}$

Direction of force is $\theta=\tan^{-1}\left ( \frac{6.414}{4.414} \right )=55.5\degree$ counter-clockwise with positive x-axis.

c)

The electric field would be the same magnitude but opposite direction as the field in part (a)

The force would be the same magnitude and direction as the force in part (b)