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Question Number 153958 by DELETED last updated on 12/Sep/21

Answered by DELETED last updated on 12/Sep/21

jawaban no 2).  F_(1,2) =k×((q_1 ×q_2 )/r^2 )    r^2 =((k×q_1 ×q_2 )/F_(1,2) ) =((9×10^9 ×3×10^(−9) ×12×10^(−9) )/(81×10^(−3) ))  =((9×3×12)/(81))×10^(−6)  m^2   =4×10^(−6)  m^2   r=(√(4×10^(−6)  )) =2×10^(−3)  m=0,2 cm//

jawabanno2).F1,2=k×q1×q2r2r2=k×q1×q2F1,2=9×109×3×109×12×10981×103=9×3×1281×106m2=4×106m2r=4×106=2×103m=0,2cm//

Answered by DELETED last updated on 12/Sep/21

Jawaban no3)    k=(1/(4πε_0 ))=9×10^9    F=k×((q_1 .q_2 )/r^2 )=(1/(4πε_0 ))×((q_1 .q_2 )/r^2 )     =9.10^9  ×((10^(−8) ×10^(−8) )/((10×10^(−2) )^2 ))      =9×10^(−6)  N=0,9 dyne //

Jawabanno3)k=14πε0=9×109F=k×q1.q2r2=14πε0×q1.q2r2=9.109×108×108(10×102)2=9×106N=0,9dyne//

Answered by DELETED last updated on 12/Sep/21

Jawaban no 4  jarak q_(1,2) =x cm  jarak q_(2,3) =(18−×) cm  q_1 =+4 C  q_2 =+16 C  q_3 =−2 →agar F_R =0  maka F_(1,3) =F_(2,3)   k.((q_(1.) q_3 )/r_(1,3) ) = k. ((q_(2.) q_3 )/r_(2,3) )   ((q_(1.) q_3 )/r_(1,3) ) = ((q_(2.) q_3 )/r_(2,3) )   ((4_. (−2))/x^2 ) = (((_. −2)×16)/((18−x)^2 ))   (4/x^2 ) = ((16)/((18−x)^2 ))   (((18−x)^2 )/x^2 ) = ((16)/4)   (((18−x)^2 )/x^2 ) = 4  (((18−x))/x) = (√4) →(((18−x))/x) = 2  18−x=2x →3x=18  x=6 cm//

Jawabanno4jarakq1,2=xcmjarakq2,3=(18×)cmq1=+4Cq2=+16Cq3=2agarFR=0makaF1,3=F2,3k.q1.q3r1,3=k.q2.q3r2,3q1.q3r1,3=q2.q3r2,34.(2)x2=(.2)×16(18x)24x2=16(18x)2(18x)2x2=164(18x)2x2=4(18x)x=4(18x)x=218x=2x3x=18x=6cm//

Answered by DELETED last updated on 12/Sep/21

Jawaban no 5).  r_(AC) =2 m    r_(BC) =1 m  F_R =F_(AC) +F_(BC)         =k ((q_A q_C )/2^2 ) + k ((q_B q_C )/1^2 )      =9.10^9  (((4.10^(−4) .3.10^(−4) )/4)+((1.10^(−4) .3.10^(−4) )/1))      =9.10^9  (((4.10^(−4) .3.10^(−4) )/4)+((4.10^(−4) .3.10^(−4) )/4))      =9.10^9  (((24.10^(−8) )/4))=540 N//

Jawabanno5).rAC=2mrBC=1mFR=FAC+FBC=kqAqC22+kqBqC12=9.109(4.104.3.1044+1.104.3.1041)=9.109(4.104.3.1044+4.104.3.1044)=9.109(24.1084)=540N//

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