Problem: What are the x and y components of the electric field at the point ( 17 m, 1 m), due to a charge of 7 `microC at ( 5 m, 15 m)?
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Solution: The electric field at a point is the force per unit test charge, with the test charge located at the point. So we assume a unit test charge (1 C). The force on the test charge is one of repulsion, so will be in the direction from ( 5, 15) to ( 17, 1). The distance between charges is `sqrt[( 17- 5)^2+( 1- 15)^2] m = 18.43 m. The magnitude of the force is (9 x 10^9 N m^2/C^2)( 7 x 10^-6 C)(1 C)/( 18.43 m)^2 = 185.4 N. The x and y displacements from charge 7 `microC to the 1 C test charge are, respectively, 12 m and -14 m, so the x and y components are in proportion 12/ 18.43, and -14/ 18.43 to the force. The x and y forces are therefore ( 12/ 18.43)( 185.4 N) = 120.7 N and ( -14/ 18.43)( 185.4 N) = -140.8 N. Since these are the forces on a 1 C test charge, the electric field has components 120.7 N/C and -140.8 N/C in the x and y directions, respectively.
Generalized Response: We imagine charges Q and q at respective positions (x1, y1) and (x2, y2) in the plane, with q the 'test charge'. We find the force per unit charge on q.
By the usual means,we find that the magnitude of the force on either charge is | F | = k |Q q | / r^2. The force F12 exerted on charge 1 by charge 2 is equal and opposite to the force F21 exerted on charge 2 by charge 1. The x and y components of the force at (x2, y2) are therefore
F21x = F21 `(x2 - x1) / r and
F21y = `F21 (y2 - y1) / r, where r = `sqrt [ (x2 - x1)^2 + (y2 - y1)^2 ]. The electric field at (x2, y2) is the force per unit of test charge at that point: E = F21 / q. Putting all the above expressions together we obtain
Ex = k Q q / ( (x2-x1)^2 + (y2 - y1)^2) * (x2 - x1) / `sqrt( (x2-x1)^2 + (y2 - y1)^2) = k Q q * (x2 - x1) / ( (x2-x1)^2 + (y2 - y1)^2) ^ (3/2)
and
Ey
= k Q q / ( (x2-x1)^2 + (y2 - y1)^2) * (y2 - y1) / `sqrt( (x2-x1)^2 + (y2 - y1)^2)
= k Q q * (y2 - y1) / ( (x2-x1)^2 + (y2 - y1)^2) ^ (3/2).
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Explanation in terms of Figure(s), Extension
Figure description: The figure below depicts the charges Q and q at points (x1,y1) and (x2,y2). The legs and hypotenuse of the fundamental triangle are indicated (the hypotenuse is found using the Pythagorean Theorem; the legs are found by the obvious means). The force vectors F12 and F21 are depicted, assuming an attractive force resulting from opposite charges. The second figure shows the force vector F12 broken into components; the triangle so formed is similar to the upper triangle, so the same proportions apply between sides of both triangles; in particular the proportions (x2 - x1) / r and (y2 - y1) / r are used to obtain the x and y components of F21 . q is regarded as a 'test charge', so the electric field that (x2, y1) is F21 / q.
