# 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. One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.