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

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ycp191 ycp191 · Answer 1 · 2019-06-05T21:18:12+02:00

Answer: B

i tried putting explanation but its not working

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aristocles aristocles · Answer 2 · 2019-07-20T12:50:39+02:00

Answer:

B. [tex]2.8 \times 10^{-5} C[/tex]

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

[tex]E = \frac{kQ}{r^2}[/tex]

here we know that

[tex]E = 4.5 \times 10^5 N/c[/tex]

also we know that

[tex]r = 0.75 m[/tex]

now from above formula we have

[tex]4.5 \times 10^5 = \frac{(9\times 10^9)(Q)}{(0.75)^2}[/tex]

here we will have

[tex]Q = 2.8 \times 10^{-5} C[/tex]