Question

3. I wo point charges are placed at two of the corners of a triangle as shown in the figure. (Take qi= -11.3 uC, and 92 = 15.4 uC.) Find the magnitude and the direction of the electric field at the third corner of the triangle. {1.07e+07 N/C at 71.2° above the -X-axis} 10.0 cm - 20.0 cm

Answer 1

20 cm Anst het the triangle be ABC Ĉ at cy & we have to find So Ec= / Electric field due to t/E field due to ( charge a / charge 92 Ec- 1 92_ĝ+ I v2 (2 . 4a Go (n) 20 whose r is the distance of clorges from c point. 406 (2) Electric field due to a (-11.34() is along y direction & due t o a (+15:44C) is along na directiction. A

1 4260 11:3x10 (10x10²2) ² 47 1 15.4x10 (2) 40Go (20x102) 2 - 40 to 10-2 I where â & ŷ are unit rectors alone 1 x dy direction So the LEC) = √6² + tēla Jelita[(:3x104)2 + (154 x you)? IĘ L = (100) | 0192 - y 23 As I = 9x109 Nm2 cm2 ... so. El = 9x109 xl0-6F2 | 127.69 + 14.82 lecl - 9x 11.937.4 105 _ If = 10. xloº Nc IĘ 91.07x107 Net 4nto

Now Angle with Resultant a axis of the tano = P B TEī tan o = IEM . El 161 = Yunto x 113 x lot = 2.935 TEL X to 2 15.4 x 10-6 - 4810-2 члео tano: 2.935 .or. So 0=7112° above the x-axis 1o= tant 2.935
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