the fields produced by the charges in corners 1 and 3 are in opposite

# The fields produced by the charges in corners 1 and 3

This preview shows page 5 - 8 out of 8 pages.

the fields produced by the charges in corners 1 and 3 are in opposite directions. Since they have the same magnitudes, they combine to give zero resultant. The fields produced by the charges in corners 2 and 4 point in the same direction (toward corner 2). Thus, E C = E C2 + E C4 , Figure 1 where E C is the magnitude of the electric field at the center of the rectangle, and E C2 and E C4 are the magnitudes of the electric field at the center due to the charges in corners 2 and 4 respectively. Since both E C2 and E C4 have the same magnitude, we have E C = 2 E C2 . 1 2 3 4 1 4 + q + q + q - q The distance r , from any of the charges to the center of the rectangle, can be found using the Pythagorean theorem (see Figure 2 at the right): 2 2 (3.00 cm) +(5.00 cm) 5.83cm d 2 Therefore, 2.92 cm 2.92 10 m 2 d r Figure 2 The electric field at the center has a magnitude of 9 2 2 12 2 2 2 2 2 2 2 2(8.99 10 N m /C )(8.60 10 C) 2 1.81 10 N/C (2.92 10 m) C C k q E E r Figure 3 at the right shows the configuration given in text Figure 18.21 b. All four charges contribute a non-zero component to the electric field at the center of the rectangle. As discussed in Conceptual Example 11, the contribution from the charges in corners 2 and 4 point toward corner 2 and the contribution from the charges in corners 1 and 3 point toward corner 1. Notice also, the magnitudes of E 24 and E 13 are equal, and, from the first part of this problem, we know that E 24 = E 13 = 1.81 10 2 N/C Figure 3 The electric field at the center of the rectangle is the vector sum of E 24 and E 13 . The x components of E 24 and E 13 are equal in magnitude and opposite in direction; hence ( E 13 ) x ( E 24 ) x = 0 Therefore, C 13 24 13 13 ( ) ( ) 2( ) 2( )sin y y y E E E E E From Figure 2, we have that 5.00 cm 5.00 cm sin 0.858 5.83 cm d and 2 2 C 13 2 sin 2 1.81 10 N/C 0.858 3.11 10 N/C E E ______________________________________________________________________________ 42. REASONING AND SOLUTION The magnitude of the force on q 1 due to q 2 is given by Coulomb's law: 1 2 3 4 1 4 d 3.00 cm 5.00 cm 1 2 3 4 1 4 - q + q + q - q C E 13 E 24 1 2 12 2 12 k q q F r (1) The magnitude of the force on q 1 due to the electric field of the capacitor is given by 1C 1 C 1 0 F q E q (2) Equating the right hand sides of Equations (1) and (2) above gives 1 2 1 2 0 12 k q q q r Solving for r 12 gives 0 2 12 12 2 2 9 2 2 6 –2 2 = [8.85×10 C /(N m )](8.99×10 N m /C )(5.00×10 C) = = 5.53×10 m (1.30×10 C/m ) k q r  45. SSM REASONING The two charges lying on the x axis produce no net electric field at the coordinate origin. This is because they have identical charges, are located the same distance from the origin, and produce electric fields that point in opposite directions. The electric field produced by q 3 at the origin points away from the charge, or along the y direction. The electric field produced by q 4  #### You've reached the end of your free preview.

Want to read all 8 pages?

• Fall '12
• Vitkalov
• • •  