Answer:
Explanation:
To find the electric field you use the equation for an electrostatic electric field:
[tex]E=k\frac{q_1q_2}{r^2}[/tex]
r: distance in which E is calculated, from each charge
In the of a dipole you have two contributions to E:
[tex]\vec{E}=\vec{E_1}+\vec{E_2}[/tex]
where E1 is the electric field generated by the first charge and E2 by the second one.
A. (-13 cm, 0):
First you calculate the vectors E1 and E2:
[tex]E_1=(8.98*10^9)\frac{(1*10^{-6}C)}{(-0.13-0.03)^2}\hat{j}\\\\E_1=350781.25N/C\\\\E_2=-(8.98*10^9)\frac{(1*10^{-6}C)}{(-0.13+0.03)^2}\hat{j}\\\\E_2=-989000N/C[/tex]
Then, you sum both contributions:
[tex]\vec{E}=-547218.75N/C\hat{j}[/tex]
B. (-3cm, 10cm):
[tex]r_1=\sqrt{(0.06)^2+(0.1)^2}=0.116m\\\\\theta=tan^{-1}(\frac{0.06}{0.1})=30.96\°\\\\r_2=0.1m\\\\E_1=(8.98*10^9Nm^2/C)\frac{(1.6*10^{-6}C)}{(0.116m)^2}[cos(30.96\°)\hat{i}+sin(30.96\°)\hat{j}]\\\\E_1=[-915646\hat{i}-549306.42\hat{j}]N/C\\\\\theta=(90-30.96)+180=239.04\°\\\\[/tex]
the last angle is calculated again because the vector direction is measured from the +x axis.
and for the second vector:
[tex]E_2=(8.98Nm^2/C)\frac{1.6*10^{-6}C}{(0.1m)^2}\hat{j}\\\\E_2=1436800N/C\hat{j}[/tex]
the total E is:
[tex]\vec{E}=[-915646\hat{i}+887493.58\hat{j}]N/C[/tex]
