University:
Kentucky State University
Course:
PHY 212 | General Physics II
Academic year:

2021
Views:

142

Pages:

5
Author:

Arielle Armstrong

Coulomb’s Law (empirical law) Force between two point charges q1 and q2: • force is vector, directed along connecting line • magnitude:  q1q 2 F k 2 12 r12 Inverse squared law r12 is the distance between the charges 2 Nm 1 9 k  9 10  2 C 40 with 0  8.85 1012 C2 . 2 Nm k is Coulomb’s constant, ε0 is permittivity of free space. Example: a positive charge Q1 = +Q is located a distance d along the y-axis from the origin. A second positive charge Q2 = +Q is located at the origin and a negative charge Q3 = -2Q is located on the x-axis a distance 2d away from Q1. Calculate the net electrostatic force on Q1 due to the other two charges. y Q1=+Q d Q2=+Q 2d Q3=-2Q x Calculate the net electrostatic force on Q1 due to the other two charges. F  F2  F3 y  +Q  +Q  ˆ Q2 ˆ q1q2 ˆ F2  k 2 j  k jk 2 j 2 r12 d d F2 Q1=+Q d F3  F3x ˆi  F3y ˆj  F3 F3  F3 cos ˆi  F3 sin ˆj 2d  Q2=+Q Q3=-2Q sin   1 2 cos  3 2 x q1q3 q1q3 ˆ F3  k 2 cos i  k 2 sin ˆj r13 r13  +Q  -2Q  F3  k 2 2d    +Q  -2Q  1 ˆ 3ˆ i  k j 2 2 2  2d Calculate the net electrostatic force on Q1 due to the other two charges.  +Q  -2Q  F3  k 2  2d y F2 Q1=+Q d Q2=+Q  +Q  -2Q  1 ˆ 3ˆ i  k j 2 2 2  2d 2Q2 3 ˆ 2Q2 1 ˆ F3  k 2 i  k 2  j 4d 2 4d 2  F3 3 kQ2 ˆ 1 kQ2 ˆ F3  i  j 2 2 4 d 4 d 2d  Q3=-2Q x Q2 ˆ 3 kQ2 ˆ 1 kQ2 ˆ F  F2  F3  k 2 j + i  j d 4 d2 4 d2 3 kQ2 ˆ 3 kQ2 ˆ F i  j 2 2 4 d 4 d y F2 Q1=+Q d Q2=+Q F  F3 2d  Q3=-2Q x Quiz: If Q1 were free to move, what direction would its initial acceleration be? How would I calculate the acceleration?