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Find the voltage near a 14.0 cm diameter metal sphere that has 8.10 nC of excess positive charge on it.

Part 1 +

Give the equation used to find the electrical potential due to a point charge in terms of its charge (Q), the distance from the point charge (r), and Coulomb's constant (k). This assumes that the potential an infinite distance from the point charge is zero.

V =



Part 2

What is the voltage near the surface of the sphere?

1 Answer

5 votes

Answer:

The voltage near the surface of the sphere is approximately 1,030.71 volts.

Step-by-step explanation:

Part 1:

The equation used to find the electrical potential (V) due to a point charge in terms of its charge (Q), the distance from the point charge (r), and Coulomb's constant (k) is given by Coulomb's Law:

V = k * Q / r

where:

V = Electrical potential (voltage) at a distance r from the point charge

k = Coulomb's constant ≈ 8.99 x 10^9 N m^2 / C^2 (in SI units)

Q = Charge of the point charge

r = Distance from the point charge to the point where the potential is being measured

This equation assumes that the electrical potential an infinite distance from the point charge is zero.

Part 2:

To find the voltage near the surface of the sphere, we'll treat the sphere as a point charge at its center. The charge on the sphere is positive, so the potential near the surface will be positive. Given that the diameter of the sphere is 14.0 cm, the radius (r) will be half of that, i.e., r = 7.0 cm = 0.07 m.

The charge (Q) on the sphere is 8.10 nC (nanoCoulombs) = 8.10 x 10^-9 C.

Now, we can use Coulomb's Law to find the potential at the surface (V_surface):

V_surface = k * Q / r

Substitute the known values:

V_surface = (8.99 x 10^9 N m^2 / C^2) * (8.10 x 10^-9 C) / 0.07 m

Calculate:

V_surface ≈ 1,030.71 V

So, the voltage near the surface of the sphere is approximately 1,030.71 volts.

User Max Pierini
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