The potential V at a distance of 25.0 cm from a very small charged sphere is 48.0 V, with V taken to be zero at an infinite distance from the sphere. A) If the sphere is treated as a point charge, what is its charge? B) What is the potential at a distance of 75.0 cm from the sphere?
To determine the charge of the small sphere and the potential at a different distance, we can make use of the equation for potential due to a point charge:
V = k * (q / r)
where V represents the potential, k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2), q is the charge of the sphere (in Coulombs), and r is the distance from the sphere (in meters).
A) First, let's determine the charge of the sphere at a distance of 25.0 cm with a potential of 48.0 V:
Since the sphere is treated as a point charge, we can rearrange the equation to solve for q:
q = V * r / k
Substituting the given values, we have:
q = (48.0 V) * (0.25 m) / (8.99 x 10^9 Nm^2/C^2)
Calculating this will give you the charge of the sphere.
B) Next, let's find the potential at a distance of 75.0 cm from the sphere:
Using the same formula, we can rearrange it to solve for V:
V = k * (q / r)
Substituting the given values, we have:
V = (8.99 x 10^9 Nm^2/C^2) * (q / 0.75 m)
Calculating this will give you the potential at a distance of 75.0 cm from the sphere.