A thin copper plate of diameter 6 cm is charged to 8.40 nC. What is the strength of the electric field 0.1 mm above the center of the top surface plate? a. 2.47 times10^{5} N/C b. 1.68 times10^{5} N/C c. 3.36 times10^{5} N/C d. 4.20 time

The correct answer is (a) {eq}2.47 times10^{5} N/C{/eq}. To calculate the electric field strength, we can use the formula: {eq}E = \frac{kQ}{r^2} {/eq} where {eq}k {/eq} is Coulomb's constant ({eq}9 times 10^9 Nm^2/C^2{/eq}), {eq}Q {/eq} is the charge on the plate, and {eq}r {/eq} is the distance between the observation point and the center of the plate. Given: Charge on the plate: {eq}Q = 8.40 times 10^{-9} C {/eq} Diameter of the plate: {eq}d = 6 cm {/eq} Radius of the plate: {eq}r = d/2 = 3 cm = 0.03 m {/eq} Distance from the center of top surface plate to the observation point: {eq}h = 0.1 mm = 0.0001 m {/eq} The distance between the observation point and the center of the plate is the hypotenuse of the right triangle formed by the radius and the distance {eq}h {/eq}. Using Pythagoras theorem: {eq}d_{eff} = sqrt{r^2 + h^2} = sqrt{(0.03 m)^2 + (0.0001 m)^2} = 0.0300005 m {/eq} Now we can substitute the values in the formula: {eq}E = \frac{kQ}{d_{eff}^2} = \frac{(9 times 10^9 Nm^2/C^2) (8.40 times 10^{-9} C)}{(0.0300005 m)^2} = 2.47 times 10^5 N/C {/eq} Therefore, the strength of the electric field 0.1 mm above the center of the top surface of the plate is {eq}2.47 times10^{5} N/C{/eq}.