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        A charge 10-9 C is located at the origin in free space and another charge Q at (2,0,0) . If the X component of the electric field at (3,1,1) is zero, calculate the value of Q.
7 months ago

							We calculate field at (3,1,1) due to both charges.E=E1+E2   =(kQ ÷ |r1|2)* vector component +     (kq ÷ |r2|2)* vector component   =kQ/3 *(i+j+k) +k×10-9/11 *(3i+j+k)Now x component is zeroSo , 0 =kQ/3 +3k×10-9/11Q/3 = -3×10-9/11So Q= -3/11 × 10-9 C

7 months ago
							Electric field due to a point charge Q along a unit vector r^ at a distance r from it is(KQ / r2) r^. For charge at origin, the point (3,1,1) is at distance r =  = , r^ = {(3-0)i + (1-0)j + (1-0)k} / |r|  =  (3i + j + k) / Electric field due to charge 10 – 9 C [located at origin (0,0,0)] at (3,1,1)  is  (K x 10 – 9 / 11)((3i + j + k) / ),which is (K x 10 – 9 / 11) (3i + j + k). Similarly, for charge Q, the point (3,1,1) is at diatance r =  = ,r^ = {(3-2)i + (1-0)j + (1-0)k} /   =  (i + j + k) / .Electric field due to charge Q [located at (2,0,0)] at (3,1,1)  is  (K x Q/ 3)((i + j + k) / ),which is (K x Q/ 3) (i + j + k). x-component of electric field due to both the charges at (3,1,1) is(K x 10 – 9 / 11) (3i)  +  (K x Q/ 3) (i) = K [ (3 x 10 – 9 / 11) + (Q/ 3) ] iwhich is equal to zero.  K [ (3 x 10 – 9 / 11) + (Q/ 3) ]  =  0    (3 x 10 – 9 / 11) + (Q/ 3) = 0Solving, we get Q =  – (9 / 11) x 10 – 9 C.

7 months ago
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