26 - THEELECTRIC FIELD 26.1 Model The electric field is that of the two charges placed on the y-axis Visualize Please refer to Figure Ex26.1 We

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THE ELECTRIC FIELD 26.1. Model: Visualize: both the charges are positive, their electric fields at P are directed away from the charges. Solve: The electric field from q1 is The electric field is that of the two charges placed on the y-axis. Please refer to Figure Ex26.1. We denote the upper charge by q1 and the lower charge by q2. Because (9.0 x lo9 N m2 / Cz)(l x C) ebelow +x-axis (cos & - sin e 3) (0.05 m)’ + (0.05 m)’ Because tan6 = 5 cm/5 cm = 1, the angle 6 = 45”. Hence, zl =(1800N/C) (A;-$) -’ Similarly, the electric field from q2 is ,!?,,,,, = El + E, = 2(18OO N / C) - i^ = 25502 N / C (4 Thus, the strength of the electric field is 2550 N/C and its direction is horizontal. Assess: y-components of their fields will cancel when added. Because the charges are located symmetrically on either side of the x-axis and are of equal value, the 26.2. Model: Visualize: Please refer to Figure Ex26.2. The electric field of the positive charge q, at P is away from ql. On the other hand, the electric field of the negative charge q2 at P is toward These two electric fields are then added vertically to obtain the net electric field at P. Solve: The electric field from The electric field is that of the two charges located on the y-axis. cos & - sin 6) (9.0 x lo9 N mz / C’)(l.O x C) El= 1 = [ (0.05 m)’ + (0.05 m)’ - ( -- “I1, @below +x-axis 4m0 52 Because tan0 = 5 cm/5 cm, 8 = 45”. So, (45 +Li-L;) Jz Similarly, the electric field from * E,,,,, = + E2 = 2(1800 N / C) Thus, the strength of the electric field is 2550 N/C and its direction is vertically downward. Assess: cancel. A quick visualization of the components of the two electric fields shows that the horizontal components 26-1
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26-2 Chapter 26 26.3. Model: The electric field is that of the two 1 nC charges located on the y-axis. Visualize: Please refer to Figure Ex26.3. We denote the top 1 nC charge by q1 and the bottom 1 nC charge by q2 The electric fields ( E, and E,) of both the positive charges are directed away from their respective charges. With vector addition, they yield the net electric field Erie, at the point P indicated by the dot. Solve: The electric fields from q1 and are (9.0 x lo9 N m' / C')(1 x lo4 C) (0.05 m)' along +x-axis i =36001N/C 1 @above +x-axis Because tan@ = 10 cm/5 cm, 8 = tan-'(2) = 63.43'. So, - (9.0 x lo9 N m' / C')(I x C) E? = (cos63.43' 1 + sin63.43'33) = (3221 + 6443) N / C (0.10 m)' + (0.05 m)' The net electric field is thus E,, a,P = E, + i2 = (39222 + 6443) N / C To find the angle this net vector makes with the x axis, we calculate * @= 9.3' 644N/C 3922 N / C tan@ = Thus, the strength of the electric field at P is = J(3922 N / C)' + (644 N / C)' = 3975 N/C and makes an angle of 9.3'above the +x-axis. Assess: special symmetry relative to the charges, we expected the net field to be at an angle relative to the x-axis. 26.4. Model: Visualize: Please refer to Figure Ex26.4. We denote the positive charge by q1 and the negative charge by q2. The electric field E, of the positive charge q, is directed away from ql, but the field 2, is toward the negative charge We will add El and Solve: Because of the inverse square dependence on distance, E?
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This note was uploaded on 04/07/2008 for the course PHSX 211/212 taught by Professor Medvedev during the Spring '08 term at Kansas.

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26 - THEELECTRIC FIELD 26.1 Model The electric field is that of the two charges placed on the y-axis Visualize Please refer to Figure Ex26.1 We

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