Question

A hollow Spherical conductor of radius 12 cm is

charged to 6x10^-6 c. Find the electric field
Strength at the surface of sphere inside the sphere
at 8 cm and at distance 15m from the sphere.​

Answer 1

To find the electric field strength at various points relative to a charged hollow spherical conductor, Gauss's Law is applied. The electric field on the surface is calculated using E = kq/r², it is zero inside the sphere, and decreases with the square of the distance outside the sphere.

Step-by-step explanation:

The question pertains to finding the electric field strength of a charged hollow spherical conductor at various distances from its center. By applying Gauss's Law, we can find the electric field at the surface, inside, and outside the sphere.

For the electric field at the surface of the sphere (radius 12 cm), we use the formula E = kq/r², where k is Coulomb's constant (8.99 x 10⁹ Nm²/C²), q is the charge (6 x 10⁶ C), and r is the radius (0.12 m).

Inside the sphere (at 8 cm from the center), the electric field is 0 N/C because the charge on a conductor resides on its exterior surface only. Lastly, at a distance (15 m) from the surface, we use the same formula for the electric field as on the surface but with r = 15 m. The electric field decreases with the square of the distance from the surface of the sphere.

Answer 2

Step-by-step explanation:

Radius of the charged sphere (R)=12cm

Charge (q)=6∗10^−6C

Electric field intensity (E)=?

i) On the surface

`E=(1)/(4\pix_(e0) ) (q)/(R^2) =9*10^(9) *(6*10^(-6) )/(0.12^-2) =3.75*10^6 N/C`

ii) Inside the sphere,

E=0; Because charge enclosed by Gaussian surface is zero.

iii) Outside the sphere at distance 15cm

i.e. r=12+15=27cm=0.27m

`E=(1)/(4\pi _(e) )_(0) (q)/(r^(2) ) =9*10^9*(6 \a* *10^6)/((0.27)^2) =7.41*10^5 N/C`