Question

A van de Graaff generator has a metal sphere of radius 14.2 cm. To what potential can it be charged before the electric field at its surface exceeds 2.01 x 106 N/C (which is sufficient to break down dry air and initiate a spark)?

Answer 1

I need the answer

Answer 2

solve for me to see

Answer 3

[AI-generated answer — Explain Bot]

To find the potential to which the van de Graaff generator can be charged before the electric field at its surface exceeds 2.01 x 10^6 N/C, we need to use the formula for electric field around a conducting sphere:

E = k * (Q / r^2)

Where:
E is the electric field
k is the electrostatic constant (8.99 x 10^9 N m^2/C^2)
Q is the charge on the sphere
r is the radius of the sphere

In this case, we are given the electric field (E = 2.01 x 10^6 N/C), and the radius of the sphere (r = 14.2 cm = 0.142 m).

We can rearrange the formula to solve for the charge (Q):

Q = E * r^2 / k

Substituting the given values:

Q = (2.01 x 10^6 N/C) * (0.142 m)^2 / (8.99 x 10^9 N m^2/C^2)

Calculating this expression, we find:

Q ≈ 4.01 x 10^-8 C

Therefore, the van de Graaff generator can be charged up to approximately 4.01 x 10^-8 C before the electric field at its surface exceeds 2.01 x 10^6 N/C.