Q: Two subatomic particles have a charge of 1.0 x 10-6 C, and they are located on the x-axis at coordinates ( -1.0 m, 0.0 m ) and ( 1.0 m, 0.0 m ). Please calculate the following:
- The net electric field when a positive test charge ( P ) is situated at coordinates ( 0.0 m, 1.0 m ).
- The force upon a -2.0 x 10-9C charge at point ( P ).
- The quantity of charge ( q3 ) that, when placed at the origin, cancels the previously established E-field at point ( P ).
A3: In order to balance the E-field in question, the new E-field ( E̅3 ) must be oriented downwards on the y-axis. The previous E-field orientation ( upward ) was created by two positive charges, so a particle of negative charge ( q3 ) must be placed at the origin:
Vectors have both magnitude ( scalar ) and directional components. For example, an object moving with a velocity ( v ) in the +x-direction has a value of ( +1 )( v ) = +v, but it will have a value of ( -1 )( v ) = -v if it moves in the -x-direction. Notice that the magnitude of ( v ) is equal to that of ( -v ). Furthermore, +1 = ( v / v ), and -1 = ( -v / v ). The numerator of the preceding ratios is a vector quantity ( +/- v ), and the denominator contains its associated scalar value ( | v | ). A general expression relating the scalar and displacement components of charges separated by some distance ( r ) is as follows:
u̅r = ( r̅r / r )
Distance ( r ) is a scalar quantity, but direction ( r̅r ) is a vector quantity. A distance of 30.0 m has a very different interpretation than 30.0 m south. Furthermore, the value ( u̅r ) is a unitless quantity derived by dividing any vector by its magnitude. The process of creating a unitless vector from another is called normalization. The value of ( r ) is the scalar magnitude of the distance separating ( q3 ) from ( P ). The ( r̅r ) value is derived by subtracting the ( x ) and ( y ) coordinates of ( q3 ) from the ( x ) and ( y ) coordinates of point ( P ). This technique will give rise to components of ( r̅r ) if they exist, and they are assigned a value of ( i ) if they coincide with the +x-axis, or a value of ( j ) if it coincides with the +y-axis ( and vice versa ):
r̅r = [ ( 0.0 m – 0.0 m )i + ( 1.0 m – 0.0 m )j ]
u̅r = ( r̅r / r ) = ( j / r )
Since r = 1.0 m ( by convention ), u̅r = j.
The math is letting us know that the E-field vector associated with ( r ) has its origins in the +y-axis. Unit vectors always point away from charges that give rise to E-fields towards a region of interest at a point ( P ); however, the orientation of the E-field created at ( P ) is determined by whether or not the charge creating the field is positive or negative. Remember: positive charges are E-field sources, and negative charges are E-field sinks.
An electric field ( E̅ ) tells us how much force a charged species will experience in a given region of space. The magnitude of an electric field created at point ( P ) is inversely proportional to its distance from the charge ( or charges ) that give rise to it:
E = kq/r2
The unit vector ( u̅r ) is associated with ( E ) in the following manner:
u̅r = ( E̅ / E )
E̅ = ( E )( u̅r )
In order for the net field to be zero at ( P ), E3 must equal E̅. In order for E̅ to have a positive value, we must use the absolute value of ( q3 ) in our calculations:
| E̅3 | = | E̅ |
Recall from Part 1 that E̅ = ( 9.0 x 103 /√2 N/C )( j )
E̅3 = ( k|q3| / r23 )( u̅ )
u̅r = j
E̅3 = ( k|q3| / r23 )( j )
( k|q3| / r23 )( j ) = ( 9.0 x 103 /√2 N/C )( j )
( k|q3| / r23 ) = 9.0 x 103 /√2 N/C
r2 = 1.0 m2
k|q3| = ( 9.0 x 103 /√2 N/C )( 1.0 m2 )
k = 9.0 x 109 Nm2/C2
|q3| = [ ( 9.0 x 103 /√2 N/C )( 1.0 m2 ) ]/( 9.0 x 109 Nm2/C2 )
|q3| = 1.0 x 10-6 /√2 C
q3 = – 1.0µC/√2
RENAISSANCE PHYSICS 