Question #231538

Two charges of -6.25nC and -12.5nC are placed 25cm in a line.
Determine the electric field at a point 10cm to the left of -6.25nC and
the electric field at a point 10cm to the left of the -12.5nC

Expert's answer

E1=kqr2=−9×109×12.5×10−9(35×10−2)2=−918.36N/cE_1=\frac{kq}{r^2}=-\frac{9\times10^9\times12.5\times10^{-9}}{({35\times10^{-2}})^2}=-918.36N/c

E2=kqr2=−9×109×6.5×10−9(10×10−2)2=−5850N/cE_2=\frac{kq}{r^2}=-\frac{9\times10^9\times6.5\times10^{-9}}{({10\times10^{-2}})^2}=-5850N/cE=E1+E2=−918.36−5858=−6768.36N/cE=E_1+E_2=-918.36-5858=-6768.36N/c

E1′=kqr2=−9×109×12.5×10−9(10×10−2)2=−11250N/cE'_1=\frac{kq}{r^2}=-\frac{9\times10^9\times12.5\times10^{-9}}{({10\times10^{-2}})^2}=-11250N/c

E2′=kqr2=−9×109×6.25×10−9(15×10−2)2=−2500N/cE'_2=\frac{kq}{r^2}=-\frac{9\times10^9\times6.25\times10^{-9}}{({15\times10^{-2}})^2}=-2500N/c

E′=E1′+E2′=−11250−2500=−13750N/cE'=E'_1+E'_2=-11250-2500=-13750N/c


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