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 × 10^5 newtons/coulomb. What is the magnitude of charge on this Van der Graff generator?

A. 1.7 × 10^-7 coulombs

B. 2.8 × 10^-7 coulombs

C. 3.0 × 10^-7 coulombs

D. 8.5 × 10^-7 coulombs

Answer 1

Answer: B

i tried putting explanation but its not working

Answer 2

B.
`2.8 * 10^(-5) C`

Step-by-step explanation:

As we know that the electric field due to Van de graff generator is same as that of a point charge

so it is given by

`E = (kQ)/(r^2)`

here we know that

`E = 4.5 * 10^5 N/c`

also we know that

`r = 0.75 m`

now from above formula we have

`4.5 * 10^5 = ((9* 10^9)(Q))/((0.75)^2)`

here we will have

`Q = 2.8 * 10^(-5) C`