We have a charge q = -0.6 x 10^-6 C at origin. And a vector r = (0 m, -5 m, 0 m). Find the electric field E(r) as a vector expressed in terms of i j and k.
Ok so far I got E(r) = k (0.6e-6)/(5)^2 = 216 N/C. But this is not a vector, how do I express this in terms of I j and K?
E=kq/r^2=kq*(r as a vector)/r(magnitude)^3
E=k(-6E-6)((0,-5,0)/(0,125,0)=(0,6k/25 E-6,0)
In this problem it is simple to realize E points toward the origin, (216N/C j) ( I did not check the 216), or (0,216,0)
Could you explain to me why E points towards the origin? The vector r points in the negative direction anyway (away from origin). so -j
To express the electric field E(r) as a vector in terms of i, j, and k, you need to consider the direction of the electric field.
Since the charge q is negative, the electric field will point towards the charge. In this case, it will be in the positive y-direction (j-direction) because the charge is located along the negative y-axis.
The magnitude of the electric field E(r) that you calculated correctly is 216 N/C. To express it as a vector in terms of i, j, and k, we can write it as:
E(r) = 0i + 216j + 0k
This indicates that the electric field has no component in the x-direction (i-direction), a magnitude of 216 N/C in the y-direction (j-direction), and no component in the z-direction (k-direction).
So, the electric field E(r) can be expressed as a vector: E(r) = 216j, where j represents the unit vector in the positive y-direction.