Set 51 Problem number 1

Problem

Find the magnitude of the force between the following charges and specify if the force is attractive or repulsive:

• .0007 Coulombs at ( 9 m, -3.004 m) and .0007 Coulombs, positioned at ( 5 m, 6 m) on the x-y plane.

College and University Physics Students:  Find the direction of the force exerted on the second charge by the first, and the vector describing this force.

Solution

The force on the charges depends on the magnitudes of the charges and the distance between them. 

The distance between the charges is found using the Pythagorean Theorem:

• The x and y displacements between the points are `dx = ( 5 m - 9 m) = -4.001 m and `dy = ( 6 m - -3.004 m) = 9.004 m.
• The distance between charges is therefore

• distance between charges:   r = `sqrt( ( -4.001 m)^2 + ( 9.004 m)^2 ) = 9.853 m.

The force between the charges is thus

• force between charges = k * q1 * q2 / r^2 = (9 * 10^9 N m^2 / C^2) * | .0007 Coulombs | * | .0007 Coulombs | / ( 9.853 meters)^2 = 45.42 Newtons.

The force is repulsive if the charges are like (i.e., of like signs) and attractive if unlike.

The direction of the force is parallel to the vector r12 from the second charge to the first, in the direction of this vector if the charges are like and opposite if the charges are unlike.

• The vector r12 has x and y components r12x = x2 - x1 = 5 m - 9 m = -4.001 m and r12y = y2 - y1 = 6 m - -3.004 m = 9.004 m. 
• Since r12 is a unit vector it is equal to < -4.001 , 9.004> / | <-4.001 , 9.004> | = < -4.001 , 9.004> / `sqrt( -4.001^2, 9.004^2 ) = <-.4061, .9138 >:

• r12 = < -.4061 m, .9138 m>

• The force is thus

• F12 = (9 * 10^9 N m^2 / C^2) * .0007 Coulombs * .0007 Coulombs  / ( 9.853 m)^2 * r12 = 45.42 N/m * < -.4061 m, .9138 m > = <-18.45 Newtons, 41.5 Newtons>.

Generalized Solution

In general if we have charges q1 and q2 separated by distance r, the force between the charges has magnitude

• | F |  = k | q1 q2 | / r^2, where k = 9 * 10^9 N m^2 / C^2.

This is analogous to the force F = G m1 m2 / r^2 between two masses.   However, in the case of masses the force is always attractive, whereas with charges the force is attractive if the charges are like (i.e., of the same sign) and repulsive if unlike (i.e., of opposite signs).

Using vector notation, if the charges q1 and q2 are located at respective points (x1, y1) and (x2, y2), the magnitude of the force will be

• | F | = k | q1 q2 | / r^2 = k | q1 q2 | / | r12 |^2

where  r12 is the vector displacement | r12 | = < (x2 - x1), (y2 - y1) > from charge 1 to charge 2.

The direction of the force exerted on charge 2 by charge 1 will be parallel to the vector r12, in the direction of this vector when the charges are like and opposite when unlike. 

The vector r12 has direction defined by the unit vector u12 = r12 / | r12 | (this is a vector of magnitude 1, directed from q1 toward q2).  The force exerted on charge 2 by charge 1 is thus the product of the magnitude of the force, the unit vector u12, and if the charges are unlike the factor -1 (this factor will reverse the direction of the vector u12, so that the vector is directed toward q1 from q2, corresponding to an attractive force):

• F12 = ( k q1 q2 / | r12 | ^ 2 ) * u12 = ( k q1 q2 / | r12 | ^ 2 ) * r12 / | r12 |.

The x component of this force is

• F12x = k q1 q2 / [ (x2 - x1)^2 + (y2 - y1)^2 ]   *   [ (x2 - x1) / `sqrt( (x2 - x1)^2 + (y2 - y1)^2 ) ],

and the y component is

• F12x = k q1 q2 / [ (x2 - x1)^2 + (y2 - y1)^2 ]   *   [ (x2 - x1) / `sqrt( (x2 - x1)^2 + (y2 - y1)^2 ) ].