Two free point charges +q and +4q are a distance L apart. A third charge is
placed so that the entire system is in equilibrium. Find the location,
magnitude and sign of the third charge.
kq⋅4qL2=kq⋅q0x2k\frac{q\cdot 4q}{L^2}=k\frac{q\cdot q_0}{x^2}kL2q⋅4q=kx2q⋅q0
and
kq⋅4qL2=k4q⋅q0(L−x)2k\frac{q\cdot 4q}{L^2}=k\frac{4q\cdot q_0}{(L-x)^2}kL2q⋅4q=k(L−x)24q⋅q0
We have
kq⋅q0x2=k4q⋅q0(L−x)2→3x2+2Lx−L2=0→x=L/3k\frac{q\cdot q_0}{x^2}=k\frac{4q\cdot q_0}{(L-x)^2} \to 3x^2+2Lx-L^2=0\to x=L/3kx2q⋅q0=k(L−x)24q⋅q0→3x2+2Lx−L2=0→x=L/3 to the right of the charge qqq . Answer
kq⋅4qL2=kq⋅q0(L/3)2→q0=−49qk\frac{q\cdot 4q}{L^2}=k\frac{q\cdot q_0}{(L/3)^2}\to q_0=-\frac{4}{9}qkL2q⋅4q=k(L/3)2q⋅q0→q0=−94q . Answer
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