Force betwwn two particales q 1 a n d q 2 i s g i v e n b y : q_1\ and\ q_2\ is\ given\ by:\\ q 1 an d q 2 i s g i v e n b y :
F = k q 1 q 2 r 2 F=\frac{kq_1q_2}{r^2} F = r 2 k q 1 q 2
Let q 1 b e A , q 2 b e B , q 3 b e C a n d q 4 b e D q_1\ be\ A,\ q_2\ be
\ B,\ q_3\ be\ C\ and\ q_4\ be\ D q 1 b e A , q 2 b e B , q 3 b e C an d q 4 b e D
Force on A due to B :
F A B = 9 × 1 0 9 × 150 × 47 × 1 0 − 12 4 2 = 3.97 N a l o n g − x a x i s S i m i l a r l y , F_{AB}=\frac{9\times10^9\times150\times47\times10^{-12}}{4^2}=3.97\ N\ along\ -x\ axis\\Similarly,\\ F A B = 4 2 9 × 1 0 9 × 150 × 47 × 1 0 − 12 = 3.97 N a l o n g − x a x i s S imi l a r l y ,
Force on Adue to C :
F A C = 4.4412 N F_{AC}=4.4412\ N F A C = 4.4412 N
Direction of this force is : θ 1 = t a n − 1 5 4 = \theta_1=tan^{-1}\frac{5}{4}= θ 1 = t a n − 1 4 5 = 51.3 ° c l o c k w i s e o f x a x i s 51.3\degree\ clockwise\ of\ x\ axis 51.3° c l oc k w i se o f x a x i s
Similarly,
Force on A due to D :
F A D = 1.188 N a l o n g y a x i s F_{AD}=1.188\ N\ along\ y\ axis F A D = 1.188 N a l o n g y a x i s
resultant of F A B a n d F A D i s : F_{AB}\ and\ F_{AD}\ \ is: F A B an d F A D i s :
F A B D = 3.9 7 2 + 1.18 8 2 = 4.144 N D i r e c t i o n o f t h i s r e s u l t a n t i s : F_{ABD}=\sqrt{3.97^2+1.188^2}=4.144\ N\ \\Direction\ of\ this\ resultant\ is: F A B D = 3.9 7 2 + 1.18 8 2 = 4.144 N D i rec t i o n o f t hi s res u lt an t i s :
θ 2 = t a n − 1 3.97 1.188 = 73.34 ° a n t i c l o c k w i s e o f y a x i s \theta_2=tan^{-1}\frac{3.97}{1.188}=73.34\degree\ anticlockwise\ of\ y\ axis θ 2 = t a n − 1 1.188 3.97 = 73.34° an t i c l oc k w i se o f y a x i s
Now net force on A is given by:
F n e t = 4.14 4 2 + 4.441 2 2 + 2 × 4.144 × 4.4412 cos 214.64 ° = 2.57 N F_{net}=\sqrt{4.144^2+4.4412^2+2\times4.144\times4.4412\cos214.64\degree}=2.57\ N F n e t = 4.14 4 2 + 4.441 2 2 + 2 × 4.144 × 4.4412 cos 214.64° = 2.57 N
Direction of this net force is given by ;
θ = 121.33 ° c l o c k w i s e o f F A B D O r , ( ( 90 + 73.34 ) − 121.33 ) = 42.01 ° a n t i c l o c k w i s e o f x a x i s \theta=121.33\degree\ clockwise\ of\ F_{ABD}\\Or,\\((90+73.34)-121.33)=42.01\degree\ anticlockwise\ of\ x\ axis θ = 121.33° c l oc k w i se o f F A B D O r , (( 90 + 73.34 ) − 121.33 ) = 42.01° an t i c l oc k w i se o f x a x i s
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