Factors Influencing the Poor Study Habits of Students
SAMPLE PROBLEMS: 112- Topic 1: ELECTRIC FORCES & ELECTRIC FIELDS.
1) Charges of - 6 and + 4 nC are 3 m apart. Determine the force they exert on each other, and the electric field at a point midway between them.
|Solution: This is an electrostatics problem (stationary charges). Since the charges|[pic] |
|are of opposite sign, they will attract each other. The forces are indicated in the| |
|figure. We may immediately apply Coulomb's Law to determine the strength of these | |
|forces: | |
Felec = k q1 q2 /r2 = (9 x 109)(6 x 10-9)(4 x 10-9)/32 = 24 x 10-9 = 24 nN.
|In the second part of the problem we wish to determine the electric field at point |[pic] |
|'P', halfway between the two charges. Each source charge produces its own field at | |
|'P'. Since q1 is negative, its field E1 is directed toward q1, while the field E2 | |
|is directed away from charge q2 since it is positive. | |
The magnitudes of these fields can be immediately calculated:
E1(at P) = k q1 /(r1)2 = (9 x 109)(6 x 10-9)/(1.52) = 24 N/C
E2(at P) = k q2 /(r2)2 = (9 x 109)(4 x 10-9)/(1.52) = 16 N/C
Electric fields are vector quantities and we must add as vectors. In this case the addition is simple since both fields point in the same direction. Thus:
|Enet| = |E1| + |E2| = 24 + 16 = 40 N/C. The direction of Enet is, of course,
1) Charges of - 6 and + 4 nC are 3 m apart. Determine the force they exert on each other, and the electric field at a point midway between them.
|Solution: This is an electrostatics problem (stationary charges). Since the charges|[pic] |
|are of opposite sign, they will attract each other. The forces are indicated in the| |
|figure. We may immediately apply Coulomb's Law to determine the strength of these | |
|forces: | |
Felec = k q1 q2 /r2 = (9 x 109)(6 x 10-9)(4 x 10-9)/32 = 24 x 10-9 = 24 nN.
|In the second part of the problem we wish to determine the electric field at point |[pic] |
|'P', halfway between the two charges. Each source charge produces its own field at | |
|'P'. Since q1 is negative, its field E1 is directed toward q1, while the field E2 | |
|is directed away from charge q2 since it is positive. | |
The magnitudes of these fields can be immediately calculated:
E1(at P) = k q1 /(r1)2 = (9 x 109)(6 x 10-9)/(1.52) = 24 N/C
E2(at P) = k q2 /(r2)2 = (9 x 109)(4 x 10-9)/(1.52) = 16 N/C
Electric fields are vector quantities and we must add as vectors. In this case the addition is simple since both fields point in the same direction. Thus:
|Enet| = |E1| + |E2| = 24 + 16 = 40 N/C. The direction of Enet is, of course,