Problem:
Three conducting concentric spherical shells of radius R, 2R and 3R having charges Q,-4Q and 6Q respectively. Find the magnitude of electric field at the point on the surface of 6Q charge.

Solution:

To find the magnitude of the electric field at the point on the surface of the 6Q charge, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of the medium.

Step 1: Determine the electric field contributions from each shell

- The electric field due to a conducting shell is zero inside the shell.
- The electric field due to a conducting shell is the same as if all the charge of the shell is concentrated at the center of the shell.

Thus, the electric field at the surface of the 6Q charge is only contributed by the charge of the 6Q shell.

Step 2: Apply Gauss's law to find the electric field due to the 6Q shell

- Consider a Gaussian surface in the shape of a sphere centered at the center of the 6Q shell and with a radius of 3R.
- The charge enclosed by this Gaussian surface is 6Q.
- The electric flux through this Gaussian surface is given by Φ = E * A, where E is the electric field and A is the area of the Gaussian surface.
- Using Gauss's law, we have Φ = Qenc / ε₀, where Qenc is the charge enclosed by the Gaussian surface and ε₀ is the permittivity of free space.
- Equating the two expressions for electric flux, we get E * A = Qenc / ε₀.

Step 3: Calculate the electric field magnitude

- The area of the Gaussian surface is 4π(3R)^2 = 36πR^2.
- Plugging in the values, we have E * 36πR^2 = 6Q / ε₀.
- Solving for E, we get E = (6Q / ε₀) / (36πR^2) = Q / (6ε₀πR^2).

Therefore, the magnitude of the electric field at the point on the surface of the 6Q charge is Q / (6ε₀πR^2).

Conclusion:
The magnitude of the electric field at the point on the surface of the 6Q charge is Q / (6ε₀πR^2).