Two conducting spheres with diameters of 0.400 m and 1.00 m are separated by a distance that is

Two conducting spheres with diameters of 0.400 m and 1.00 m...

Two conducting spheres with diameters of $0.400 \mathrm{m}$ and $1.00 \mathrm{m}$ are separated by a distance that is large compared with the diameters. The spheres are connected by a thin wire and are charged to $7.00 \mu \mathrm{C}$. (a) How is this total charge shared between the spheres? (Ignore any charge on the wire.) (b) What is the potential of the system of spheres when the reference potential is taken to be $V=0$ at $r=\infty ?$

Key Concepts

Electric Potential and Reference Point

Electric potential is a measure of potential energy per unit charge, and it is defined relative to a chosen reference point. For problems involving charges at large distances, it is common to set the potential at infinity to zero. This reference simplifies the interpretation of potentials for the system and is fundamental in determining the absolute potential of charged bodies.

Conservation of Charge

This principle asserts that the total charge in an isolated system remains constant despite any redistribution among its parts. When two conductors are connected, although the charge might rearrange to equalize the potential across them, the sum of the individual charges remains equal to the initially applied total charge.

Capacitance of a Spherical Conductor

The capacitance of a single isolated sphere is given by the formula C = 4???R, where R is the radius of the sphere. This relationship is used to determine how much charge a sphere can hold at a given potential. In the context of connected spheres, it helps to quantify the charge distribution based on their sizes.

Electrostatic Equilibrium in Conductors

When conductors are connected by a wire, they reach the same electric potential in electrostatic equilibrium. This means that any excess charge will redistribute among the conductors until the potential is uniform across all connected bodies. This concept is essential for determining how charge is shared between different conductors that are electrically joined.

Transcript

0:00 All right, hello.

00:01 This question asks about electric potential and charge distribution for two spheres of different radii that are connected by a long, thin wire.

00:09 So we're given the radii of these two spheres and the total charge.

00:15 And if these are connected with a conducting wire, they're going to distribute their charge such that their potentials are the same.

00:23 And so what that means for us is that the ratio of their charge is q1, or sorry, qa over qb is going to be directly proportional to ra over rb.

00:43 Now, we know that we don't know what qa is versus qb.

00:47 That's what we're trying to find out.

00:48 But we do know that our total charge is going to be the sum of our two charges, so qa plus qb.

00:55 And so solving this for, we'll solve it for qa.

00:58 It'll be our total charge minus our charge on b.

01:02 Plugging that into our equation on the left, we'll have q total minus qb over qb.

01:11 And i'll simplify that a little more here.

01:14 That'll be q total over qb minus 1.

01:21 And that equals the ratio of our radii.

01:25 Solving this then for qb, we're going to get, so after doing some algebra, we get qb is the total charge times the radius of b over the sum of the radii.

01:42 And that's just multiplying, adding the one to the right side, and then inverting everything and multiplying by qt.

01:51 And we have all of these values.

01:53 So plugging in those values from our question above will give us a qb of 2 .00 microcoulombs.

02:03 And then we know that qa, therefore, must just be 7 minus 2, which will be 5 .00 microcoulombs.

02:12 Now, part b of this question asked us to find the electrical potential as a function of r.

02:17 And in order to do this, we need to account for the fact that we have to figure out where we are in relation to the spheres.

02:24 Because if we're at a point between the spheres, the equation for the potential will be different than if we're, say, way out here on the right...

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