 |
| Numerical 04 |
|
| 04. The charges shown in the figure are stationary. Find the force on 4mC charge due to the other two. |
| |
 |
| |
| Suggested answer: |
| |
| Let F1 be the magnitude of the force on charge +4mC due to charge +2mC. |
| |
 |
| |
| Then, |
| |
 |
| |
| or F1 = 1.8N |
| |
| Again, if F2 is the magnitude of the force on charge +4mC due to charge +3mC, then |
| |
 |
| |
| F2 = 2.7 N |
| |
| Net horizontal component, Fx = F2 cos60o- F1 cos60o |
| |
| or Fx = (2.7) (0.5) - (1.8) (0.5) |
| |
| = (2.7-1.8) 0.5 N = 0.45 N towards left |
| |
| Net vertical component in the vertically upward direction, |
| |
| Fy = F2 sin60o+ F1 sin60o |
| |
| |
| |
| = ( 2.7 ) ( 0.866 ) + ( 1.8 ) (0.866 ) |
| |
| |
| |
| = (2.7+1.8) 0.866 N = 3.897 N |
| |
| |
| |
| = 3.9 N (Rounding off to two significant figures) |
| |
| Net force is given by |
| |
 |
| |
| |
| |
 |
| |
 |
| |
| = 3.9 N |
| |
| |
| |
| (Rounding off to two significant figures) |
| |
| If q is the angle which F makes with the positive direction of y-axis, then |
| |
 |
| |
