Two point charges of 4 C and 16 C are kept at a distance of 0.18 m. A third point charge of 1 C is kept between them such that the potential energy of the system is minimum. What is the magnitude of the electric field at the location of the 1 C charge?
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Let the 4 C charge be at \(x=0\) and the 16 C charge at \(x=d=0.18\ m\), with the 1 C charge at position \(x\) between them. The potential energy of the system is \(U(x)=k\dfrac{(4)(16)}{d}+k\dfrac{(4)(1)}{x}+k\dfrac{(16)(1)}{d-x}\). For minimum \(U\), \(\dfrac{dU}{dx}=-k\dfrac{4}{x^{2}}+k\dfrac{16}{(d-x)^{2}}=0\), which gives \(\dfrac{4}{x^{2}}=\dfrac{16}{(d-x)^{2}}\), i.e. the net force on the 1 C charge is zero at this point (so \(x=0.06\ m\), \(d-x=0.12\ m\)). Since force equals charge times electric field, a zero net force at the location of the 1 C charge means the net electric field there, due to the 4 C and 16 C charges, is also zero. The correct option is (a).
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