Question

A charge q1 = 5 μC, is at the origin. A second charge q2 = -3 μC is on the x-axis 0.8 m from the origin. The electric field at a point on the y-axis 0.5 m from the origin is:

Answer 1

Answer:

$\vec{E}=\vec{E_1}+\vec{E_2}=[25856\hat{i}+163443.2\hat{j}]N/C$

Step-by-step explanation:

The electric field is given by:

$E=k\frac{q_1q_2}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

at the point P(0,0.5m) you have both x ad y component of the electric field. For the particle q1 you have:

$\vec{E_1}=Ex\hat{i}+Ey\hat{j}\\\\\vec{E_1}=0\hat{i}+(8.89*10^9Nm^2/C^2)\frac{(5*10^{-6}C)}{(0.5m)^2}\hat{j}=179600N/C\hat{j}$

for the particle q2, it is necessary to compute the angle between the E vector and the axis, by using the distance y and x. Furthermore it is necessary to know the distance from q2 to the point P.

$\vec{E_2}=Excos\theta \hat{i}-Eysin\theta \hat{j}\\\\\theta=tan^{-1}(\frac{0.5}{0.8})=32\°\\\\r=\sqrt{0.5^2+0.8^2}=0.94m\\\\\vec{E_2}=(8.89*10^9Nm^2/C^2)\frac{(-3*10^{-6C})}{(0.94m)^2}[cos(32)\hat{i}-sin(32)\hat{j}]\\\\=[25856.06\hat{i}-16156.71\hat{j}]N/C$

Finally, by adding E1 and E2 you obtain:

$\vec{E}=\vec{E_1}+\vec{E_2}=[25856\hat{i}+163443.2\hat{j}]N/C$