Question

A gaussian surface contains two charged spherical conductors A and B having radii 3mm and 2mm respectively if the respective surface charge densities are 10 microcoulomb per metre squared minus 5 microcoulomb per metre square the total normal electric induction over the Asian surface will be​

Answer 1

Explanation:

The charge of magnitude +q present inside the spherical shell induces a charge of magnitude −q 0n the inner part of the shell. Therefore, for the charge on the inner part, the surface charge density will be due to −q.

It is given by:

σ

1

=

surface area of inner part

Totalcharge

⇒

4πr

1

2

−q

The charge on the outer shell will be +q and the surface charge density at the outer part will be due to sum of q and Q charges present on the outer shell.

σ

2

=

Outersurfacearea

Totalcharge

=

4πr

2

2

Q+q

...(ii)

(b) If we consider a loop inside an irregular conductor, then will be no work done on it as there is no field present inside the cavity. So, there will be no field inside the cavity even if it is irregular in shape.

Answer 2

Given,

A Gaussian surface has two charged spherical conductors with radii of 3mm and 2mm, respectively.

A has a surface charge density of 10 μC/m²

B has a surface charge density of -5 μC/m²

To find,

Total enclosed charge

Solution,

We can simply solve the numerical issue using the approach outlined below.

We know that;

∴ σ = $\frac{Q}{A}$

⇒ Q = σ A

       = σ.4πR²

Thus,

Q₁ = σ₁ A₁

    = 10 * 10⁻⁶ * 4π * 9 * 10⁻⁶ C

Q₂ = σ₂ A₂

     = -5 * 10⁻⁶ * 4π * 4 * 10⁻⁶ C

Now,

The total charge enclosed = Q₁ - Q₂

                                             = 10⁻¹² (90-20) 4π

                                             = 280 * π * 10⁻¹²

                                             = 8.8 * 10⁻¹⁰ C

Thus, the total charge enclosed is 8.8 * 10⁻¹⁰ C.