# Factors Influencing the Poor Study Habits of Students

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• Published : February 18, 2013

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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, toward the negative charge.

2). The strength of the electric field in the region between two parallel deflecting plates of a cathode-ray oscilloscope is 25 kN/C. Determine the force exerted on an electron passing between these plates. What acceleration will the electron experience in this region?

|Solution: As indicated in the figure, one of the plates is positively charged, while the other is|[pic] | |negatively charged. The electric field between parallel charged plates is constant (as long as we| | |stay in the middle away from the edges of the plates), and is directed from the + plate toward | | |the - plate. The electric force on a charge q placed in a field E is: | | |Fon q = q E . | | | | | |Since q is negative for an electron, then the force on the electron is opposite to the direction| | |of E, i.e., to the left (toward the + plate). The magnitude of this force is: | | | | | ||Fon e-| = e E = (1.6 x 10-19)(25 x 103) = 4 x 10-15 N . | |

While this is an extremely small force, we find the acceleration is quite large: a = F/m = (4 x 10-15)/(9.11 x 10-31) = 4.4 x 1015 m/sec2.

The direction of the acceleration is, of course, the...