Question

At 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 × 10⁵ newtons/coulomb. what is the magnitude of charge on this van der graff generator? a. 1.7 × 10⁻⁵ coulombs b. 2.8 × 10⁻⁵ coulombs c. 3.0 × 10⁻⁵ coulombs d. 8.5 × 10⁻⁵ coulombs

Answer 1

Step-by-step explanation:

To find the magnitude of the charge on the van der Graff generator, you can use the formula for electric field due to a point charge:E=k∗∣q∣r2E=r2k∗∣q∣​Where:EEis the electric field strength (given as 4.5×1054.5×105N/C).kkis the electrostatic constant (8.99×1098.99×109 N m²/C²).qqis the charge we want to find.rris the distance from the charge to the point where the electric field is measured (given as 0.75 meters).We can rearrange the formula to solve for qq:q=E∗r2kq=kE∗r2​Now, plug in the values:q=(4.5×105 N/C)∗(0.75 m)28.99×109 N m²/C²q=8.99×109N m²/C²(4.5×105N/C)∗(0.75m)2​Calculating this gives:q≈2.8×10−5 coulombsq≈2.8×10−5coulombsSo, the magnitude of the charge on the van der Graff generator is approximately 2.8 × 10⁻⁵ coulombs, which corresponds to option b.