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Electric Field

Read from Lesson 4 of the Static Electricity chapter at The Physics Classroom:





MOP Connection: Static Electricity: sublevels 10 and 11

1. The standard metric units of measurements for electric field strength are Newton/Coulomb (N/C).

2. The direction of the electric field vector is defined as __the direction that a positive test charge

would be pushed or pulled if placed at the given location.

Use the electric field equations to answer the following questions.

3. A test charge of +1.0x10-6 C experiences a force of 0.050 N. The electric field strength is _5.0 x 104 N/C_.

4. A test charge of +1.0x10-6 C experiences a force of 0.10 N. The electric field strength is _1.0 x 105 N/C_.

5. An object with a charge of 2.0x10-4 C creates an electric field. A test charge of +1x10-6 C experiences a force of 0.050 N. The electric field strength is _5.0 x 104 N/C_.

6. An object with a charge of 2.0x10-4 C creates an electric field. A test charge of +2x10-6 C experiences a force of 0.10 N. The electric field strength is _5.0 x 104 N/C_.

7. An object with a charge of 4.0x10-4 C creates an electric field. A test charge of +1x10-6 C experiences a force of 0.10 N. The electric field strength is _1.0 x 105 N/C_.

|8. An object with a charge of Q creates an electric field. A positive test charge, q, is used to test the|[pic] |

|strength of the field. Use this scenario to answer the following questions: | |

|a. If the charge of the test charge q is doubled, then it will experience | |

__2X__ (2X, 4X, 1/2, 1/4-th, the same) force; the electric field strength at this location will be __the same as__ (2X, 4X, 1/2, 1/4-th, the same as) the original value.

b. If the charge of the object Q is doubled, then the test charge will experience __2X__ (2X, 4X, 1/2, 1/4-th, the same) force; the electric field strength at this location will be __2X__ (2X, 4X, 1/2, 1/4-th, the same as) the original value.

c. If the distance between the charge and the test charge is doubled, then the test charge will experience __1/4-th the__ (2X, 4X, 1/2, 1/4-th, the same) force; the electric field strength at this location will be __1/4-th__ (2X, 4X, 1/2, 1/4-th, the same as) the original value.

9. Use your understanding of electric force and electric field to fill in the following table.

| |Charge creating the E |Charge used to test the |Force experienced by |Electric Field | |

| |field |E field |test charge (N) |Intensity |Distance |

| |(C) |(C) | |(N/C) |(fictional units) |

|a. |4.0 x10-4 C |1.0 x 10-6 C |0.20 N |2.0x105 N/C |d |

|b. |4.0 x10-4 C |2.0 x 10-6 C |0.40 N |2.0 x105 N/C |d |

|c. |8.0 x10-4 C |1.0 x 10-6 C |0.40 N |4.0x105 N/C |d |

|d. |8.0 x10-4 C |2.0 x 10-6 C |0.80 N |4.0 x105 N/C |d |

|e. |8.0 x10-4 C |1.5 x 10-6 C |0.60 N |4.0x105 N/C |d |

|f. |8.0 x10-4 C |1.0 x 10-6 C |0.10 N |1.0 x105 N/C |2d |

|g. |8.0 x10-4 C |2.0 x 10-6 C |0.20 N |1.0x105 N/C |2d |

|h. |8.0 x10-4 C |1.0 x 10-6 C |0.10 N |1.0x105 N/C |2d |

|i. |4.0 x10-4 C |Answers vary. |Answers vary. |8.0 x105 N/C |0.5 d |

|j. |4.0 x10-4 C |Answers vary. |Answers vary. |8.0x105 N/C |0.5 d |

In the above table, concepts and mathematical computations are blended to arrive at the answers. The concept of great importance is that the electric field intensity (column 4) depends upon the quantity of charge on the source (column 1) that creates the electric field and the distance from the source (column 5). These two variables are the only variables that affect the strength of the electric field. If neither of these two quantities is changed, then the electric field intensity will not change. By the time one progresses to the bottom of the table (row j), this concept must be used to determine the electric field intensity. In row j, columns 1 and 5 contain identical values as in row i. Therefore, the electric field intensity in row j is the same as in row i.

The mathematical computations involve relating columns 2-4 of the table to one another by the equation E = F/q where E is the electric field intensity, q is the test charge and F is the force experienced by the test charge. If any two of these three variables are known, then the third variable can be calculated. By the time one progresses to the bottom of the table (rows I and j), the answers will vary since only one of the three variables are known. For these two rows, any answers that satisfy the E=F/q equation could be considered correct.

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