Question

Match the information given in column-I with that in column-II and mark the correct code given below.

Column I		Column II
A	For an infinitely long line charge, electric field at a distance $r$ from the line is	i	inversely proportional to $r^2$ when $r<R/2$
B	For a uniformly charged non-conducting sphere, electric field at a distance $r$ from centre is $(R=$ radius of sphere)	ii	radially outward and inversely proportional to $r$
C	Electric field at a distance $r$ from the centre of a spherical charged conductor of radius. $R$, with a concentric cavity of radius $R/2$, is	iii	inversely proportional to $r^2$ when $r>R$
D	A spherical charged conductor of radius $R$ with a cavity of radius $R/2$ has a point charge $q(q>0)$ at its centre. Electric field at a distance $r$ from the centre is	iv	increases with $r$, reaches a maximum and then decreases

• A. i - (q) , ii - (s) , iii - (r) , iv - (p)
• B. i - (p) , ii - (q) , iii - (r) , iv - (s)
• C. i - (q) , ii - (r) , iii - (s) , iv - (p)
• D. i - (r) , ii - (s) , ii - (p) , iv - (q)

Answer 1

Answer: (A)

(i) In the symmetry, we except the field to be directed radially outward to depend on the perpendicular distance $r$ from the wire. In the cylindrical symmetry, the field will be same at all points on a Gaussian surface that is a cylinder with the wire along its axis.
In figure, $E$ is perpendicular to this surface at all points. For Gauss's law, we need a closed surface, so we include the flat ends of cylinder. Since $E$ is parallel to the ends, there is no flux through the ends (the cosine of the angle between $E$ and $dA$ on the end is $\left(\cos 90^{\circ }-0\right)$
For Gaussian surface, Gauss's law gives
$\oint E\cdot dA=E(2\pi rl)=\frac{Q_{\text{enclosed }}}{{\epsilon }_0}=\frac{\lambda l}{{\epsilon }_0}$
Where $l$ is the length of our chosen Gaussian surface (l < < length of wire) and $2\pi r$ is its circumference. Hence
$E=\frac{1}{2\pi {\epsilon }_0}\cdot \frac{\lambda }{r}$
(ii) Since, the charge is distributed symmetrically in the sphere, the electric field at all points must again be symmetric. $E$ depends only $r$ and is directed radially outward (or inward if $\theta <0$ ).
(iii) For the region $r>R,E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$
(iv) A Gaussian surface of radius $r<R/2$ only encloses the centre charge $q$. The electric field will therefore, be the field of the single charge.
$E(r<R/2)=\frac{q}{4\pi {\epsilon }_0{r}^2}$