Question

A charge of 5.97mC is at the point (x, y) - (1.70m, 2.80m). What is the electric field vector at the point (x,y) (-3.90m, -4.60m) due to this charge? show the direction please and show in X, Y axis as well, pretty please. explanation!

Answer 1

Given:

Charge, Q=5.97mC

Charge located at point q=(q_x,q_,y)=(1.70 m, 2.80 m)

Electric field to be calculated at point e=(e_x,e_,y)=(-3.90 m, -4.60 m)

Electric field=E=?

Solution:

Distance between q and e= r

Calculating r:

Now we know that for a given charge Q the Electric field at a point r distance away is given by:

and

Calculating E:

pointing radially outward along with the line joining points q and e.

The slope of the line:

The angle that this line makes with x-axis:

The angle that Electric field makes with x-axis:

Now,

Note: Here θ is the angle that E makes with the negative x-axis.

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