Two conducting spheres are connected by a thin wire and are in electrostatic equilibrium. The diameter of the smaller sphere is 3.0 cm and the diameter of the larger one is 8.0 cm. If the smaller sphere is found to be charged to 2.0 nC, what is the charge of the larger sphere?

To solve this problem, we need to use the principle of charge conservation, which states that the total charge before and after any interaction remains the same.

Let's assume the charge on the larger sphere is Q. Since the smaller sphere is charged to 2.0 nC, we can write the equation as:

2.0 nC + Q = Q

By rearranging the equation, we find:

Q - Q = -2.0 nC

-2Q = -2.0 nC

Now, we can solve for Q:

Q = (-2.0 nC) / (-2)

Q = 1.0 nC

Therefore, the charge on the larger sphere is 1.0 nC.

To find the charge on the larger sphere, we can use the principle of charge conservation. In electrostatic equilibrium, the total charge of an isolated system remains constant.

The charge on the smaller sphere is given as 2.0 nC, which is equal to 2.0 × 10^(-9) C. Let's denote the charge on the larger sphere as Q.

Since the spheres are connected by a thin wire, they are in electrical contact. So, the total charge on both spheres combined is equal to the sum of the charges on each individual sphere.

Q_small + Q_large = Total charge on both spheres

Substituting the values we know:

2.0 × 10^(-9) C + Q_large = Total charge on both spheres

Now, we need to find the total charge on both spheres. Since both spheres are in electrostatic equilibrium, the charges distribute themselves uniformly on the surface. We can assume that the spheres are perfect conductors, and the charges reside on the outer surface of the sphere.

The charge on a conductor is given by the product of its surface charge density and its surface area.

The surface area of a sphere is given by:

Surface Area = 4πr²

Where r is the radius of the sphere.

Let's calculate the surface area of both spheres:

For the smaller sphere:
Radius = diameter / 2 = 3.0 cm / 2 = 1.5 cm = 0.015 m
Surface Area_small = 4π(0.015 m)²

For the larger sphere:
Radius = diameter / 2 = 8.0 cm / 2 = 4.0 cm = 0.04 m
Surface Area_large = 4π(0.04 m)²

Now, let's find the total charge on both spheres:

Total charge on both spheres = Surface Charge Density × Surface Area_small + Surface Charge Density × Surface Area_large

We can divide both sides of the equation by Surface Charge Density to obtain:

Total charge on both spheres / Surface Charge Density = Surface Area_small + Surface Area_large

Now, we need to find the ratio of the surface areas:

Surface Area_large / Surface Area_small = (4π(0.04 m)²) / (4π(0.015 m)²)

Simplifying the equation:

Surface Area_large / Surface Area_small = (0.04 m / 0.015 m)²

Surface Area_large / Surface Area_small = (8/3)²

Surface Area_large / Surface Area_small = 64/9

So, the total charge on both spheres / Surface Charge Density = (64/9) + 1 (Adding 1 because the smaller sphere also has a charge)

Now, we can substitute the total charge on both spheres / Surface Charge Density from our previous equation into the charge conservation equation:

2.0 × 10^(-9) C + Q_large = (64/9) + 1

Simplifying the equation:

Q_large = (64/9) + 1 - 2.0 × 10^(-9) C

Finally, we can solve for Q_large to find the charge on the larger sphere.