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 × 10ˆ-7 coulombs
B. 2.8 × 10ˆ-7 coulombs
C. 3.0 × 10ˆ-7 coulombs
D. 8.5 × 10ˆ-7 coulombs

Answer 1

Step-by-step explanation:

As we know from the formula of electric field as

E = \frac{kq}{r^2}

here we will have

k = 8.99 \times 10^9

r = 0.75 m

E = 4.5 \times 10^5 N/c

now we will have

4.5 \times 10^5 = \frac{8.99 \times 10^9 q}{0.75^2}

from above equation we have

q = \frac{4.5 \times 10^5 (0.75)^2}{8.99 \times 10^9}

now we have

q = 2.8 \times 10^{-5}

Answer 2

As we know from the formula of electric field as

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

here we will have

`k = 8.99 * 10^9`

`r = 0.75 m`

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

now we will have

`4.5 * 10^5 = (8.99 * 10^9 q)/(0.75^2)`

from above equation we have

`q = (4.5 * 10^5 (0.75)^2)/(8.99 * 10^9)`

now we have

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