Respuesta :
Given:
The charge at the origin is,
[tex]\begin{gathered} Q_1=97.32\text{ }\mu C \\ =97.32\times10^{-6}\text{ C} \end{gathered}[/tex]The second charge is,
[tex]\begin{gathered} Q_2=14.5\text{ }\mu C \\ =14.5\times10^{-6}\text{ C} \end{gathered}[/tex]The second charge is at a distance of
[tex]\begin{gathered} r=11.88\text{ cm} \\ =0.1188\text{ m} \end{gathered}[/tex]from the origin
To find:
The work to place the second charge
Explanation:
The work to place the second charge is,
[tex]\begin{gathered} W=\frac{kQ_1Q_2}{r} \\ Here,\text{ k=9}\times10^9\text{ N.m}^2.C^{-2} \end{gathered}[/tex]substituting the values we get,
[tex]\begin{gathered} W=\frac{9\times10^9\times97.32\times10^{-6}\times14.5\times10^{-6}}{0.1188} \\ =106.9\text{ J} \end{gathered}[/tex]Hence, the required work is 106.9 J.
