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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?

1 Answer

6 votes

Given

Charge is


q_1=7.22\text{ nC}

The second charge is


q_2=-2.6\text{ nC}

The distance between the charges,


r=4.46\text{ cm=4.46}*10^(-2)m

To find

The electric field strength

Step-by-step explanation

The electric field strength is given by


F=k(q_1q_2)/(r^2)

Putting the values,


\begin{gathered} F=9*10^9(7.22*10^(-9)*2.6*10^(-9))/((4.46*10^(-2))^2) \\ \Rightarrow F=8.49*10^(-5)N \end{gathered}

Conclusion

The electric field strength is


8.49*10^(-5)N

User Nidhish Krishnan
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