Answer to Question #209124 in Electric Circuits for Lorie

Question

Two point charges, QA = +18 μC and QB= -5 μC, are separated by a distance r= 10 cm. What is the magnitude of the electric force. The constant k= 8.988 x 109 Nm2C

Answer 1

From Coulomb's law, we know that the magnitude of the electric force will be equal to

$F=k\frac{|q_Aq_B|}{r^2}$

Then, we substitute (q_A=18 $\mu$C, q_B=-5 $\mu$C, r=10 cm and k = 8.988 X 10⁹ Nm²/C²) and we can calculate the magnitude of such force:

$F=(8.988\times10^9\,Nm^2/C^2)\large{\frac{|(18\mu C)(-5\mu C)|}{(10\,cm)^2}}$

$\implies F=\frac{(8.988\times10^9\,Nm^2/C^2)|(18\times10^{-6} C)(-5\times10^{-6} C)|}{(0.1\,m)^2}$

$\implies F=\frac{(8.988)(18)(5)}{\large{(10^{-1})^2}} \times10^{9-12}\,N$

$\implies F=80892 \times10^{-3}\,N=80.892 \,N$

In conclusion, the magnitude of the electric force between these two charges under the present conditions is approximately equal to 80.892 N.

Reference:

• Serway, R. A., & Jewett, J. W. (2018). Physics for scientists and engineers. Cengage learning.