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Answers   E1x, E1y = -2.56 x 107 N/C ,0   E2x , E2y = 1.04 x 107  N/C ,0    E3x, E3y = 5.76 x 107  N/C, 0   F3x , F3y = 2.20 x 105 , 1.34 x 105   N Electric field at point A, 5.0cm Left of P,{:[E=E_(1)-E_(2)],[=(kq_(1))/(r_(1)^(2))-(kq_(2))/(r_(2)^(2))],[q_(1)=q_(2)=q" (say) "],[:.quad E=kq[(1)/(r_(1)^(2))-(1)/(r_(2)^(2))]],[=9xx10^(9)xx2xx10^(-6)[(1)/((2.5 xx10^(-2))^(2))-(1)/((7.5 xx10^(-2))^(2)]):}],[=18 xx10^(3)[(1)/(6.25 xx10^(-4))-(1)/(56.25 xx10^(-4))]H//c],[=(18 xx10^(3))/(10^(-4))[(1)/(6.25)-(1)/(56.25)]N//c],[=18 xx10^(7)[0.16-0.018]N//c],[=18 xx10^(7)xx0.142],[=2.56 xx10^(7)N//c]:}Part BPoint A is 5cm. directly above P. Electric field will be as shown.At point A, Components E_(1)sin theta and E_(2)sin theta are equal and opposite, so they will cancel out. Het electric field at A,{:[E=E_(1)cos theta+E_(2)cos theta],[=((kq_(1))/(r^(2))+(kq_(2))/(r_(2)))cos theta],[" But "|q_(1)|=|q_(2)|" and "],[r_(1)=r_(2)],[:.quadE^(˙)=(2kq)/(r^(2))cos theta]:}{:[:.E=(2kq)/(r^(2))(AP)/(r)=(2kqAP)/(r^(3))],[" How, "AP=5xx10^(-2)cm],[r=sqrt((2.5 xx10^(-2))^(2)+(5xx10^(-2))^(2))],[r=5.6 ... See the full answer