# Determine the charge of the sphere if the potential at a point located 50 cm from

**Determine the charge of the sphere if the potential at a point located 50 cm from the surface of the sphere is 4 V. The radius of the sphere is 5 cm.**

We translate all the values from given to the SI system:

l = 50 cm. = 0.5 m.

R = 5 cm. = 0.05 m.

Potential of a field point located at a distance from r center of the sphere:

φ = q / (4π * ε0 * r), where q is the magnitude of the sphere’s charge, r is the distance from the center of the sphere to the point of the field, ε0 is the electrical constant ε0 = 8.85 * 10 ^ -12 F / m.

Distance from the center of the sphere to the point of the field with potential φ:

r = R + l.

Substitute in the expression to determine the potential:

φ = q / (4π * ε0 * r) = q / (4π * ε0 * (R + l)).

Let us express from this expression – the charge:

q = φ * (4π * ε0 * (R + l).

Substitute the numerical values and determine the charge of the sphere:

q = φ * (4π * ε0 * (R + l) = 4 * (4 * π * 8.85 * 10 ^ -12 * (0.5 + 0.05) = 2.44 * 10 ^ -10 Cl …

Answer: the charge of the sphere is 2.44 * 10 ^ -10 C.