Question

a metallic sphere of radius 5 cm has a charge of 6 micro coulomb. The surface charge density is

Answer 1

We can find the area of the sphere as we know the radius
After that we can apply the formula

Sigma = charge/Area
(Surface charge density)

We will get the ans 7.64 x 10^-4

Ur welcome ✌

Answer 2

Answer:

The surface charge density of the sphere is 1.91×10⁻⁴Cm⁻².

Explanation:

The charge density is the measurement of the accumulation of the electric charge in a given particular field. It measures the amount of electric charge per unit area. i.e.

$S_D=\frac{Q}{A}$           (1)

$S_D$=surface charge density

Q=charge on the surface

A=area of the surface

From question,

The radius of the sphere=5cm=5×10⁻²m

Charge on the sphere(Q)=6μC=6×⁻⁶C

Area of the sphere =4πr²

So,

$A=4\pi \times (5\times 10^-^2)^2=4\times 3.14\times 25\times 10^-^4=3.14\times 10^-^2 m^2$       (2)

By substituting the value of Q and A in equation (1) we get;

$S_D=\frac{6\times 10^-^6}{3.14\times 10^-^2}=1.91\times 10^-^4 Cm^-^2$

Hence, the surface charge density of the sphere is 1.91×10⁻⁴Cm⁻².