Give few electric flux examples with explaination

Q) What is the flux due to electric field E = 3 × 103\(hat{i},N{{C}^{-1}}\) through side of the square 10cm, when it is held normal to E?

Answer: According to the question, E = 3 × 103i NC–1. Side of square (S) = 10 cm = 0.1 m.

Area of square (A) = (side)2 = (0.1)2 = 1 × 10–2 m2

Hence, electric flux through the square,\(phi =E.A=(3times {{10}^{3}}){{.10}^{-2}}hat{i}=30,N{{m}^{2}}{{C}^{-1}}\) [1]

Q) Given a uniform electric field E = 5 × 103\(hat{i},N{{C}^{-1}}\), find the flux of this field through a square of 10 cm on a side whose plane is parallel to the YZ-plane. If plane of the square rotated to make an angle 30o with X axis what would be the flux through the same square?

Answer: Given, electric field intensity \(E=5times {{10}^{3}}hat{i}N{{C}^{-1}}\)

Magnitude of electric field intensity |E| = 5 × 103 NC–1

Side of square, S = 10 cm = 0.1 m

Area of square, A = (0.1)2 = 0.01 m2

The plane of the square is parallel to the YZ-plane. Hence, the angle between the unit vector normal to the plane and electric field is zero. [1]

i.e. Ɵ = 0°

∴Flux through the plane,

\(phi =|E|times Acos theta Rightarrow phi =5times {{10}^{3}}times 0.01,cos ,0{}^circ\)= \(phi =50,N{{m}^{2}}{{C}^{-1}}\)

If the plane makes an angle of 30° with the X-axis, then Ɵ = 60°

∴Flux through the plane,

\(phi =|E|times Atimes cos ,60{}^circ\)

= 5 × 103 × 0.01 × cos 60° = 25 Nm2C–1[1]

Q) A sphere S1 of radius R1 encloses a net charge Q. If there is another concentric sphere S2 of radius R2(R2 > R1) enclosing charge 2Q, find the ratio of the electric flux through S1 and S2.If medium inside the sphere(S2) is replaced by a medium with dielectric constant K. How does the flux through the sphere S2 changes?

Answer: According to Gauss’ law,

Flux through\({{S}_{1}},{{phi }_{1}}=frac{Q}{{{varepsilon }_{0}}}\) …….. [i]

Flux through \({{S}_{2}},{{phi }_{2}}=frac{Q+2Q}{{{varepsilon }_{0}}},=frac{3Q}{{{varepsilon }_{0}}}\) ……. (ii)

On dividing Equation (i) by Equation. (ii), we get

\({{phi }_{1}}/{{phi }_{2}}=1/3\) [1]

There is no change in the flux through S1 with dielectric medium inside the sphere S2. [1]

Q) Two concentric metallic spherical shells of radii R and 2R are given charge \({{q}_{1}}\) and \({{q}_{2}}\) respectively. Charge density of the shells are same. Determine the ratio \({{q}_{1}}\) : \({{q}_{2}}\)

Answer:Surface charge density, \(sigma =frac{Q}{4pi {{R}^{2}}}\)

According to the question, surface charge density, s = constant

Let \({{q}_{1}}\) and \({{q}_{2}}\) are two charges

Hence,

Charge \({{q}_{1}}\) = 4πR2s ………………. (i)

Charge \({{q}_{2}}\) = 4π(2R)2s ………………… (ii)

On dividing equation (i) with Equation (ii),

\(therefore ,,,,,,,,,,,,,,frac{{{q}_{1}}}{{{q}_{2}}}=frac{4pi {{R}^{2}}sigma }{4pi {{(2R)}^{2}}sigma }=frac{1}{4}\)

Q) A thin straight infinitely long conducting wire having charge density λ is enclosed by a cylindrical surface of radius r and length l, its axis coinciding with the length of the wire.Find the flux passing through the cylindrical surface.

Answer:

A thin straight conducting wire will be a uniform linear charge distribution.

Let q amount of charge is enclosed by the cylindrical surface.

∴Linear charge density,

\(lambda =frac{q}{l} \ therefore ,,,,,,,,,q=lambda l,,,,,,,,,,,,,,,,,,……,(i),,,,,,,[1] \\)

By Gauss’ theorem,

∴Total electric flux through the surface of cylinder

\(phi =frac{q}{{{varepsilon }_{0}}},,,,,,,,,,,,,,,,,,,,,,,,,,[Gauss’theorem] \ phi =frac{lambda l}{{{varepsilon }_{0}}},,,,,,,,,,,,,,,,,,,,,,,,,,[From,Eq.(i)] \\)

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