Question

Question 2 2.5 pts Three charges Q1 = -14.2nC, Q2 = Q3 = +8.0nC are placed on the vertices of a square of side 2cm as shown in the figure below. Find the direction of the electric field at the empty vertex. Give the direction with respect to positive x axis. (Do not round up intermediate values. Only round your final answer to the nearest whole number)

Answer 1

Solution = -14.2nc g=+8.one a=2cm= 2x10²m IBDI= a X-axes ( 45 ) Eos Oz=+8.onc E02,03 The dectric field at point 'D' due to charges. Og = +8.0mc = +8.0x109c and 03= +8.0X Toc is given by Eogh Eo a 2 = LITE (2cm) 2 - Eor - LTE * 12x152 m2 = 4TTE (2x152)2 = 1 8.0xlo =) Fog = 1 8.0x109 4TIE (2x152) since, charge og = + 8.0x109 c is positive. therefore, the dectric field at point 'D' would be away from the charge & 2 = +8.0x10 9 cire. Log = 1 8 would be directed along CD 4T8 2x102 as shown in the above figure. 3.0x109

Similarly the dectric fied at point 'D' due to change Q3= + 8.0x109 c would be directed along Ao as shown in the above figure in e Eoz= 0 = 8.0x10 9 - 4TTE. (2x152) 2 Eoz = t 8.0x109 along A D 103 - UTIE 12x1022 47Eo (a)2 Now, since the deetric field on and Eo3 are perpendicelor to each other therefore their resultent Eog,03 is given by E02,03 = I Eogh+ 1E0312 along BD Ez = 12 Eogl along BD { since | E03] = 1E03] Fog, 03=52 L 8.0x10 g along RB TIE (2x15 232 along BD Eog, og = 52x gx10 °x 8.0x10' along BD $ Since, (2x102)2 LITIES 9x16 - Eos, 03 = 25.45 x104 ton along BB - (1)

Now, since the charge = - 14.2 C = - 14.2 x 109 c is negative. Therefore, the clectic field at point D' would be directed to words DB directon i. c. towards the charge o. Therefore, we have E = LITE OD OR = 476 14 2 2 9 12 Eoi = LITE 14 : 2 X 1 X 102 2 along Do Eoi = 9x19 x 142x109 along DB - 2 (2x102)2 Foi = 15.975 x10 " X along DB - (2)

Now, from equation (1) and (2), we have E09,03 ) Eoi since, Eo2,03 and Eo, are in opposite direction Therefore, the resultant dectric field Ek at point 'D' would be along Bi direction as shown in above direction. The ER is given as ER = 0,03 – Ei ER = 25.45 x104 15.975 x104 = 9.475 x104 → ER = g. 475x104 to along BD = ER ~ 9.5x104 to along Bo ER = 9.5x104 to along 3 The direction of dectric field 'ER' with respect to positive x-axis is 315° in the counter clock-wise and 45° in the clock-wise direction as shown in the above figure, Therefore, the direction with respect to x-axis is 315ºm coum tey clock-wise direction or 45° in clock-wise direction