Question

A point charged particle of 12 μC is held at the center of a conducting spherical shell of inner radius 6 cm and outer radius 10 cm. The shell has total charge 18 μC . What is the magnitude of the electric field.

Answer 1

The magnitude of the electric field at a point outside a conducting spherical shell is given by E = 162 x 10² N/C / r², where E is the electric field and r is the distance from the center of the shell.

Step-by-step explanation:

To find the electric field at a point outside a conducting spherical shell, we can use Gauss's law:

1. Draw a Gaussian surface in the shape of a sphere that encloses the charge inside the conducting shell.
2. The electric field at all points on the Gaussian surface is perpendicular to the surface.
3. The electric field is constant in magnitude and direction on the Gaussian surface.
4. Use Gauss's law to relate the electric field on the Gaussian surface to the charge enclosed by the surface.

In this case, the point charge inside the conducting shell is 12 μC and the charge on the shell is 18 μC. Since the shell is conducting, the electric field inside the shell is zero. Therefore, the electric field at all points outside the shell is the same as the electric field at the surface of the shell, which is given by:

E = k * Q / r²

Where E is the electric field, k is the electrostatic constant (9 x 10^9 N m²/C²), Q is the charge on the shell (18 μC), and r is the distance from the center of the shell.

Substituting the values into the formula, we get:

E = (9 x 10^9 N m²/C²) * (18 x 10^-6 C) / r²

Simplifying the equation, we get:

E = 162 x 10² N/C / r²

Therefore, the magnitude of the electric field at a point outside the conducting shell is given by E = 162 x 10² N/C / r².