Question #200008

q1=+10 nC and q2=-5 nC are separated by 0.10 m dinstance. If a third charge q3=2 nC is placed along with the first two charges, where should it be so that the net electric force of q1 and q2 on q3 is zero (0)?


1
Expert's answer
2021-06-01T14:35:30-0400


F23=kq2q3r2F_{23}=\dfrac{kq_2q_3}{r^{2}}


F13=kq1q3r2F_{13}=\dfrac{kq_1q_3}{r^{2}}


let distance between q2 and q3 be x.q_2\ and \ q_3\ be \ x.



so the force of the charge by q2q_2 on q3\ q_3 is denoted by F23F_{23} and the force on charge q1q_1 on q3q_3 is denoted by F13F_{13} .


Distance between q3 and q1q_3\ and \ q_1 is given 0.10+x0.10+x


Distance between q1 and q2q_1\ and \ q_2 is given by 0.10 meters


now equation both the forces we get -


F13=F23F_{13}=F_{23}



kq1q3(0.10+x)2=kq2q3x2\dfrac{kq_1q_3}{(0.10+x)^{2}}=\dfrac{kq_2q_3}{x^{2}}


== 0.5\sqrt{0.5} =xx+0.10=\dfrac{x}{x+0.10}


=0.70=xx+0.10=0.70=\dfrac{x}{x+0.10}


=0.70x+0.70×0.10=x0.70x+0.70\times 0.10=x


=0.70×0.10=x0.70x=0.70\times 0.10=x-0.70x


=0.07=0.30x=0.07=0.30x


=x=0.23m=x=0.23m


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS