Two metallic spheres have radii of 10 cm and 5 cm,respectively.The electric field strength on the surface of each sphere is 3600 N/C.The two spheres are then connected together by a long,thin metal wire.Determine the electric field strength on the surface of 5-cm radius sphere while they are connected.

Question

Quizplus · Accepted Answer

When two conductive spheres are connected by a wire, they share the same potential (voltage). The electric field (E) on the surface of a sphere is given by $E = \frac{kQ}{r^2}$, where $Q$ is the charge on the sphere, $r$ is the radius of the sphere, and $k$ is Coulomb's constant. Since the potential $V = \frac{kQ}{r}$ is the same for both spheres when connected, and the spheres initially have the same electric field strength, the charges will redistribute in such a way that the potential remains constant across both spheres. Given that the electric field strength on the smaller sphere (radius = 5 cm) needs to be found after connection, and knowing that the potential across both spheres is the same, we can deduce that the charge will redistribute in proportion to their radii. The electric field on the smaller sphere will increase because it will hold more charge per unit area (since $E \propto \frac{Q}{r^2}$ and $Q$ is now distributed based on the same potential across different radii).Since the larger sphere has twice the radius of the smaller sphere, it initially had four times the surface area (since surface area $A \propto r^2$). If the electric field strength was the same before connection, the total charge was distributed according to their surface areas. Upon connection, the charge redistributes to maintain equal potential, not equal electric field strength. The smaller sphere, with less surface area, ends up with a higher electric field strength due to a higher concentration of charge per unit area.Given the initial electric field strength is 3600 N/C for both, and considering the relationship between the radii and how charge redistributes, the electric field strength on the smaller sphere will increase. Without needing to calculate specific charges or potentials, we understand conceptually that the electric field strength must increase on the smaller sphere due to the conservation of charge and the principle of equal potentials in connected conductors. Among the given options, only one suggests an increase from the initial 3600 N/C, which is 6000 N/C.

Two Metallic Spheres Have Radii of 10 Cm and 5