Determine the magnitude of the electric field 14.8 cm directly above an isolated 3.68times 10^-5 C charge. Express your answer with the appropriate units. E=Value Units

To determine the magnitude of the electric field directly above an isolated charge, we use the formula for the electric field due to a point charge:<br /><br />\[ E = \frac{k \cdot |q|}{r^2} \]<br /><br />where:<br />- \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),<br />- \( q \) is the magnitude of the charge (\( 3.68 \times 10^{-5} \, \text{C} \)),<br />- \( r \) is the distance from the charge to the point where the electric field is being calculated (14.8 cm = 0.148 m).<br /><br />Plugging in the values:<br /><br />\[ E = \frac{(8.99 \times 10^9) \cdot (3.68 \times 10^{-5})}{(0.148)^2} \]<br /><br />First, calculate the numerator:<br /><br />\[ 8.99 \times 10^9 \times 3.68 \times 10^{-5} = 3.30792 \times 10^5 \]<br /><br />Next, calculate the denominator:<br /><br />\[ (0.148)^2 = 0.021904 \]<br /><br />Now, divide the numerator by the denominator:<br /><br />\[ E = \frac{3.30792 \times 10^5}{0.021904} \approx 1.51 \times 10^6 \, \text{N/C} \]<br /><br />So, the magnitude of the electric field directly above the charge is:<br /><br />\[ E = 1.51 \times 10^6 \, \text{N/C} \]