Respuesta :
Given:
The charge
[tex]\begin{gathered} q1\text{ = 200 }\mu C \\ =200\times10^{-6}\text{ C} \end{gathered}[/tex]is at a distance x1 = 20 cm = 0.2 m.
The charge
[tex]\begin{gathered} q2\text{ = -300 }\mu C \\ =-300\times10^{-6}\text{ C} \end{gathered}[/tex]is at a distance x2 = 30 cm = 0.3 m.
The charge
[tex]\begin{gathered} q3\text{ = -400 }\mu C \\ =-400\times10^{-6}\text{ C} \end{gathered}[/tex]is at a distance x3 = 40 cm = 0.4 m.
To find the potential at x = 0
Explanation:
The potential can be calculated by the formula
[tex]V=\frac{kq}{x}[/tex]Here k is Coulomb's constant whose value is 9 x 10^(9) Nm^2/C^2
Potential due to charge 1 is
[tex]\begin{gathered} V1\text{ =}\frac{kq1}{x1} \\ =\frac{9\times10^9\times200\times10^{-6}}{0.2} \\ =9\times10^6\text{ V} \end{gathered}[/tex]Potential due to charge 2 is
[tex]\begin{gathered} V2\text{ =}\frac{kq2}{x2} \\ =\frac{9\times10^9\times-300\times10^{-6}}{0.3} \\ =-9\times10^6\text{ V} \end{gathered}[/tex]Potential due to charge 3 is
[tex]\begin{gathered} V3\text{ =}\frac{kq3}{x3} \\ =\frac{9\times10^9\times-400\times10^{-6}}{0.4} \\ =-9\times10^6\text{ V} \end{gathered}[/tex]Thus, the net potential at x = 0 is
[tex]\begin{gathered} V_{net}\text{ = V1+V2+V3} \\ 9\times10^6-9\times10^6-9\times10^6 \\ =-9\times10^6\text{ V} \end{gathered}[/tex]