91Ó°ÊÓ

To minimize the potential at \(x=0\), we need to arrange the charges in a symmetrical manner. We can put the charges in the following order: -1.0 µC, +1.0 µC, -1.0 µC, +1.0 µC By doing this, the net potential at \(x=0\) will be minimized, as the positive and negative charges will cancel each other out at \(x=0\).

We are given that the three charges are +1.0 µC, and the far-right charge is -1.0 µC. So, the arrangement is: +1.0 µC, +1.0 µC, +1.0 µC, -1.0 µC We can use the electric potential formula to calculate the potential at the origin for each charge separately and then add the results to find the total potential. For the first charge: \(V_1 = \dfrac{k (1.0 \times 10^{-6}C)}{1.0m} = \dfrac{8.99 \times 10^9 N m^2 C^{-2} \times 10^{-6}C}{1.0m}\) For the second charge: \(V_2 = \dfrac{k (1.0 \times 10^{-6}C)}{2.0m} = \dfrac{8.99 \times 10^9 N m^2 C^{-2} \times 10^{-6}C}{2.0m}\) For the third charge: \(V_3 = \dfrac{k (1.0 \times 10^{-6}C)}{3.0m} = \dfrac{8.99 \times 10^9 N m^2 C^{-2} \times 10^{-6}C}{3.0m}\) For the fourth charge (negative): \(V_4 = \dfrac{k (-1.0 \times 10^{-6}C)}{4.0m} = \dfrac{8.99 \times 10^9 N m^2 C^{-2} \times (-10^{-6}C)}{4.0m}\) Now, we can find the total potential at the origin by adding the potentials from all four charges: \(V_{total} = V_1 + V_2 + V_3 + V_4\) \(V_{total} = \dfrac{8.99 \times 10^9 N m^2 C^{-2} \times [(10^{-6}C) + (10^{-6}C)/2 + (10^{-6}C)/3 - (10^{-6}C)/4]}{1.0m}\) After calculating, we find: \(V_{total} = 4.23 \times 10^3 V\) Thus, the potential at the origin in this charge arrangement is \(4.23 \times 10^3 V\).