select the correct answer.\nat a distance of 0.75 meters from its center, a van der graff generator…

select the correct answer.\nat a distance of 0.75 meters from its center, a van der graff generator interacts as if it were a point - charge, with that charge concentrated at its center. a test charge at that distance experiences an electric field of 4.5×10^5 newtons/coulomb. what is the magnitude of charge on this van der graff generator?\na. 1.7×10^(-5) coulombs\nb. 2.8×10^(-5) coulombs\nc. 3.0×10^(-5) coulombs\nd. 8.5×10^(-5) coulombs

select the correct answer.\nat a distance of 0.75 meters from its center, a van der graff generator interacts as if it were a point - charge, with that charge concentrated at its center. a test charge at that distance experiences an electric field of 4.5×10^5 newtons/coulomb. what is the magnitude of charge on this van der graff generator?\na. 1.7×10^(-5) coulombs\nb. 2.8×10^(-5) coulombs\nc. 3.0×10^(-5) coulombs\nd. 8.5×10^(-5) coulombs

Answer

Explanation:

Step1: Recall electric - field formula

The formula for the electric field due to a point - charge is $E = \frac{kQ}{r^{2}}$, where $E$ is the electric field, $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$ is the Coulomb's constant, $Q$ is the charge, and $r$ is the distance from the charge. We need to solve for $Q$.

Step2: Rearrange the formula for $Q$

From $E=\frac{kQ}{r^{2}}$, we can get $Q=\frac{Er^{2}}{k}$.

Step3: Substitute the given values

Given $E = 4.5\times 10^{5}\ N/C$, $r = 0.75\ m$, and $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$. $Q=\frac{(4.5\times 10^{5}\ N/C)\times(0.75\ m)^{2}}{9\times10^{9}\ N\cdot m^{2}/C^{2}}$ First, calculate $(0.75)^{2}=0.5625$. Then, $(4.5\times 10^{5})\times0.5625 = 2.53125\times 10^{5}$. $Q=\frac{2.53125\times 10^{5}}{9\times10^{9}}= 2.8125\times10^{-5}\ C\approx2.8\times 10^{-5}\ C$

Answer:

B. $2.8\times 10^{-5}$ coulombs