Set 51 Problem number 8

Problem

A charge of 18 `microC is located at ( 21 m,-23.01 m).  

• If the electric field due to this charge is observed at the point ( 13 m,-17.01 m), what will be its magnitude and direction?

Solution

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

We calculate the components of the electric field vector as usual:

• Since the charge is positive, the force on the 1 C test charge is one of repulsion, so will be in the direction from ( 21,-23.01) to ( 13,-17.01).
• The distance between charges is `sqrt[( 13- 21)^2+(-17.01--23.01)^2] m = 10 m.
• The magnitude of the force is (9 x 10^9 N m^2/C^2)( 18 x 10^-6 C)(1 C)/( 10 m)^2 = 1620 N.
• The x and y displacements from charge 18 `microC to the 1 C test charge are, respectively, -8.001 m and 6 m, so the x and y components are in proportion -8.001/ 10, and 6/ 10 to the force.
• The x and y forces are therefore (-8.001/ 10)( 1620 N) = -1296.324 N and ( 6/ 10)( 1620 N) = 972.0001 N.
• Since these are the forces on a 1 C test charge, the electric field has components -1296.324 N/C and 972.0001 N/C in the x and y directions, respectively.
• We note that since the x component is negative and the y component positive, the vector is in the second quadrant.

We then use the standard procedure to calculate the magnitude and direction of the field vector:

• The resultant force therefore has magnitude `sqrt[(-1296.324 N/C)^2+( 972.0001 N/C)^2] = 1620 N/C.
• Its direction is arctan( 6/-8.001) + 180 deg = 143.1 degrees, where the 180 deg has been added to the arctan to place the vector in the second quadrant.

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 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 angle of the force is either

• aTan( (y2 - y1) / (x2 - x1)) or
• aTan( (y2 - y1) / (x2 - x1)) + 180 deg,

depending on whether F is repulsive (F positive) or attractive (F negative).

Explanation in terms of Figure(s), Extension

The figure below depicts the charges Q and q as attracting charges 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 part of the figure shows the force vectors F21 and F12. F12 makes angle aTan( (y2 - y1) / (x2 - x1) ) with the x direction.  

• This angle might or might not place the force vector in the correct quadrant.  If not, we add 180 deg to the value of the arcTangent in order to place the vector in the correct quadrant (this is necessary when, and only when, the force has a negative x component).

Figure(s)