Question

a point charge is placed at the common centre of two concentric spheres.what is the ratio of electric flux associated with the surface if r2=3r1​

Answer 1

Answer:

E4.1 Electric Flux

In Chapter 2 we showed that the strength of an electric field is proportional to the number

of field lines per area. The number of electric field lines that penetrates a given surface is

called an “electric flux,” which we denote as ΦE . The electric field can therefore be

thought of as the number of lines per unit area.

Figure 4.1.1 Electric field lines passing through a surface of area A.

Consider the surface shown in Figure 4.1.1. Let A = Anˆ r be defined as the area vector

having a magnitude of the area of the surface, , and pointing in the normal direction,

. If the surface is placed in a uniform electric field E

A

nˆ ur

that points in the same direction

as nˆ , i.e., perpendicular to the surface A, the flux through the surface is

ˆ Φ = E E A⋅ = E⋅n A = EA r r r (4.1.1)

On the other hand, if the electric fieldE

ur makes an angle θ with (Figure 4.1.2), the

electric flux becomes

nˆ

n cos Φ = E E A⋅ = EA θ = E A r r (4.1.2)

where En = E⋅nˆ is the component of E r r

perpendicular to the surface.

Figure 4.1.2 Electric field lines passing through a surface of area A whose normal makes

an angle θ with the field.xplanation:

Answer 2

Answer: ratio of the flux associated with the surface is 1:1

Explanation:
Let the amount of charge kept at the centre of the spheres = q Coulomb

Amount of charge enclosed by inner sphere = q

therefore by Gauss's theorem flux associated with the inner sphere = q/εo----------------------1

Amount of charge enclosed by outer sphere = q

therefore by Gauss's theorem flux associated with the outer sphere = q/εo----------------------2

Therefore ratio of the flux associated with the two surfaces will be by 1 and 2

q/εo/q/εo = 1: 1
#SPJ3