Question: Q. 6. (i) Define electric flux. Write its S.I. unit.

(ii) A small metal sphere carrying charge +Q is located at the centre of a spherical cavity inside a large uncharged metallic spherical shell as shown in the figure. Use Gauss’s law to find the expressions for the electric field at points P1 and P2.

(iii) Draw the pattern of electric field lines in this arrangement. [Delhi, O.D. Comptt. I, II, III, 2012]

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Solution:

Ans. (i) Try yourself, Similar to Q. 2, Very Short Answer Type Questions

(ii) Try yourself, Similar to Q. 12, Short Answer Type Questions-II (iii) The direction of electric field is shown in the figure.

[CBSE Marking Scheme, 2012]

[IIQ.7.(i) An electric dipole of dipole moment p consists of point charges +q and q separated by a distance 2a apart. Deduce the expression for the electric field E due to the dipole at a distance x from the centre of the dipole on its axial line in terms of the dipole moment p. Hence show that in the limit x»a,E2p(4πε0x3).

(ii) Given the electric field in the region E=2xi, find the net electric flux through the cube and the charge enclosed by it. R&U [Delhi I, II, III 2015]

Ans. (i)

Electric field intensity at point P due to charge q,

$$ \vec{E}{-q}=\frac{1}{4 \pi \varepsilon{0}} \cdot \frac{-q}{(x+a)^{2}}(\hat{x}) $$

Due to charge +q,

$$ \vec{E}{+q}=\frac{1}{4 \pi \varepsilon{0}} \cdot \frac{q}{(x-a)^{2}}(\hat{x}) $$

Net electric field at point P,

$$ \begin{aligned} & \vec{E}=\vec{E}{-q}+\vec{E}{+q}^{+q} \ & \vec{E}=\frac{q}{4 \pi \varepsilon_{0}} \times\left(\frac{1}{(x-a)^{2}}-\frac{1}{(x+a)^{2}}\right)(\hat{x}) 1 / 2 \end{aligned} $$

E=q4πε0×(4aqx(x2a2)2)(x^)

=q4πε0×(q×2a)2x(x2a2)2(x^) =q4πε0×2px(x2a2)2(x^)

1/2

FFor x»a,(x2a2)2x4

E=q4πε0×2px3(x^)=14πε02px3

(ii) From the given diagram,

Only the face perpendicular to the direction of x-axis, contribute to the electric flux. The remaining faces of the cube give zero contribution.

Total flux ϕ=ϕI+ϕII

=IEds+IIEds1/2 =0+2(a)a21/2 =2a31/2

Charge enclosed, q=ε0ϕ=2a3ε0

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