Set 51 Problem number 6

Problem

A charge of 1 `microC is located at (-19.01 m,-15.01 m).  

• If the electric field due to this charge is observed at the point ( 17 m, 15 m), what will be its x and y components?

Solution

The electric field at a point is the force per unit test charge, with the test charge located at the point.

We can assume a unit test charge--i.e., a charge of 1 C.

• Since the 1 Coulomb charge and the assumed test charge are both positive, the force on the test charge is one of repulsion, so will be in the direction from (-19.01,-15.01) to ( 17, 15).
• The distance between charges is `sqrt[( 17--19.01)^2+( 15--15.01)^2] m = 46.87 m.
• The magnitude of the force is (9 x 10^9 N m^2/C^2)( 1 x 10^-6 C)(1 C)/( 46.87 m)^2 = 4.096 N.
• The x and y displacements from charge 1 `microC to the 1 C test charge are, respectively, 36.01 m and 30.01 m, so the x and y components are in proportion 36.01/ 46.87, and 30.01/ 46.87 to the force.
• The x and y forces are therefore ( 36.01/ 46.87)( 4.096 N) = 3.146 N and ( 30.01/ 46.87)( 4.096 N) = 2.622 N.
• Since these are the forces on a 1 C test charge, the electric field has components 3.146 N/C and 2.622 N/C in the x and y directions, respectively.

Generalized Solution

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 0(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).

Explanation in terms of Figure(s), Extension

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 at (x2, y1) is F21 / q.

Figure(s)