Two spherical conductors are connected by a wire. Each carry positive charges $Q_A$ and $Q_B$. Conductor A has a larger radius than conductor B. How do the electric field $E_A$ and $E_B$ at the conductor surfaces compare?
Added by Albert L.
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Step 1: The electric field at the surface of a conductor is given by: $$E = \frac{\sigma}{\epsilon_0}$$ where $\sigma$ is the surface charge density and $\epsilon_0$ is the permittivity of free space. Show more…
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Two conducting spheres are initially not connected. Sphere A has a charge of +2q and radius r0, Sphere B has a charge of +4q and radius 2r0. a. Find the surface charge η for each sphere. Which one has the larger charge per unit area? b. Now the two spheres are connected by a very thin conducing wire. What is the charge per area on each sphere? What is the total charge on each sphere? You may assume the surface area of the wire is so small compared to the spheres that it can be ignored. c. What is the electric field at the surface of each sphere? d. What is the electric potential at the surface of each sphere? Did you calculate the answer you expected? Explain.
Adi S.
1wo spherical conductors $\mathrm{A}$ and $\mathrm{B}$ of radii $\mathrm{lmm}$ and $2 \mathrm{~mm}$ are separated by a distance of $5 \mathrm{~mm}$ and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of sphere of $\mathrm{A}$ and $\mathrm{B}$ is (A) $1: 2$ (B) $2: 1$ (C) $4: 1$ (D) $1: 4$
Two spherical conductors $A$ and $B$ of radii $1 \mathrm{~mm}$ and $\mathrm{mm}$ are separated by a distance of $5 \mathrm{~cm}$ and are uniformly charged. If the spheres are connected by conducting wire, then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres $A$ and $B$ is (A) $4: 1$ (B) $1: 2$ (C) $2: 1$ (D) $1: 4$
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