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

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

Explanation:

The electric field is given by:

`E=k(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)((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)((0.5)/(0.8))=32\°\\\\r=√(0.5^2+0.8^2)=0.94m\\\\\vec{E_2}=(8.89*10^9Nm^2/C^2)((-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`