Introduction:
In this problem, we need to determine the maximum number of electrons that can be placed on a conducting sphere of radius 1m in order to avoid electrical breakdown, given the breakdown electric field strength in air.

Understanding the Problem:
To understand the problem, we need to have knowledge of the concept of electrical breakdown and its relationship with electric field strength. Electrical breakdown occurs when the electric field in a medium exceeds a certain threshold value, causing the medium to become conductive. In air, the breakdown electric field strength is typically around 3*10^6V/m.

Approach:
To calculate the maximum number of electrons that can be placed on the conducting sphere, we need to determine the charge that can be placed on the sphere without exceeding the breakdown electric field strength.

1. Electric Field Inside a Conducting Sphere:
When excess charge is placed on a conducting sphere, it distributes itself uniformly on the surface, creating an electric field inside the sphere. The electric field inside a conducting sphere is zero.

2. Electric Field at the Surface of a Conducting Sphere:
The electric field at the surface of a conducting sphere can be calculated using the formula:
E = σ/ε₀,
where E is the electric field, σ is the surface charge density, and ε₀ is the permittivity of free space.

3. Surface Charge Density:
The surface charge density can be calculated using the formula:
σ = Q/A,
where σ is the surface charge density, Q is the charge on the conducting sphere, and A is the surface area of the sphere.

4. Maximum Charge on the Sphere:
To avoid electrical breakdown, the electric field at the surface of the conducting sphere should not exceed the breakdown electric field strength. Therefore, we can write the equation:
E ≤ E_breakdown,
which can be rewritten as:
Q/(4πε₀r²) ≤ E_breakdown,
where r is the radius of the sphere.

Solving the Problem:
1. Calculating the Surface Area of the Sphere:
The surface area of a sphere can be calculated using the formula:
A = 4πr².

2. Rearranging the Equation:
By rearranging the equation Q/(4πε₀r²) ≤ E_breakdown, we can solve for Q:
Q ≤ 4πε₀r²E_breakdown.

3. Substituting the Values:
Substituting the given values of r = 1m and E_breakdown = 3*10^6V/m into the equation, we get:
Q ≤ 4πε₀(1m)²(3*10^6V/m).

4. Calculating the Value:
Using the value of ε₀ = 8.854*10^-12 C²/Nm², we can calculate the maximum charge on the sphere:
Q ≤ 4π(8.854*10^-12 C²/Nm²)(1m)²(3*10^6V/m).

5. Calculating the Maximum Number of Electrons:
The maximum charge on the sphere can be expressed in terms of the elementary charge (e):
Q = Ne,
where N is the number of electrons and e is the elementary charge.

Therefore, we can write the equation:
Ne ≤ 4π(8.854*10^-12 C²/Nm²)(1m)²(3*10