A + 7.22 nC point charge and a - 2.6 nC point charge are 4.46 cm apart. What is the electric field strength at the midpoint between the two charges?

Respuesta :

Given

Charge is

[tex]q_1=7.22\text{ nC}[/tex]

The second charge is

[tex]q_2=-2.6\text{ nC}[/tex]

The distance between the charges,

[tex]r=4.46\text{ cm=4.46}\times10^{-2}m[/tex]

To find

The electric field strength

Explanation

The electric field strength is given by

[tex]F=k\frac{q_1q_2}{r^2}[/tex]

Putting the values,

[tex]\begin{gathered} F=9\times10^9\frac{7.22\times10^{-9}\times2.6\times10^{-9}}{(4.46\times10^{-2})^2} \\ \Rightarrow F=8.49\times10^{-5}N \end{gathered}[/tex]

Conclusion

The electric field strength is

[tex]8.49\times10^{-5}N[/tex]