A non-conducting ring of radius $0.5m$ carries a total charge of $1.11 \times {10^{ - 10}}C$ distributed non-uniformly $\int\limits_{l = \infty }^{l = 0} {( - E \cdot dl)}$ ($l=0$ being centre of ring) in volt is?
(A) +2
(B) -1
(C) -2
(D) Zero

Hint
Here we will be using the formulas, $E.dl = - dv$ (where, $E$= electric field, $dl$= small length of non conducting ring and $V$= potential ) and $\int\limits_{l = \infty }^{l = 0} {(dv) = } {V_{centre}} - {V_\infty }(V_{center}$ because $l=0$).

Complete step by step answer
Given, total charge carries by a non-conducting ring$= 1.11 \times {10^{ - 10}}C$.
Also, given the radius of the non-conducting ring $R=0.5m$.
So, we have to find the value of $\int\limits_{l = \infty }^{l = 0} {( - E \cdot dl)}$ ;
$
\Rightarrow \int\limits_{l = \infty }^{l = 0} {( - E \cdot dl)} = \int\limits_{l = \infty }^{l = 0} {(dv)} \{ \because - E.dl = dv\} \\
   \Rightarrow \int\limits_{l = \infty }^{l = 0} {( - E \cdot dl)} = {V_{centre}} - {V_\infty }\{ \because \int\limits_{l = \infty }^{l = 0} {(dv) = } {V_{centre}} - {V_\infty }\} \\
   \\
 $
$\Rightarrow {V_{centre}} = \dfrac{{9 \times {{10}^9}(1.11 \times {{10}^{ - 10}})}}{{0.5}}{}
 \Rightarrow {V_{centre}} = 2V$
And, ${V_\infty } = 0 (\because$ potential at infinity is $0)$
So,
$
   \Rightarrow \int\limits_{l = \infty }^{l = 0} {( - E \cdot dl)} = {V_{centre}} - {V_\infty } \\
   \Rightarrow \int\limits_{l = \infty }^{l = 0} {( - E \cdot dl)} = 2V - 0V \\
   \Rightarrow \int\limits_{l = \infty }^{l = 0} {( - E \cdot dl)} = 2V \\
$
Therefore, the value of line integral is $\int\limits_{l = \infty }^{l = 0} {( - E \cdot dl)} = 2V$.
Option (A) is correct.

Note
The electric potential of a material is the potential difference in potential energy per unit charge between two point charges in an electric field. $V = k\dfrac{q}{r}$ {where, k= coulomb constant, q=charge and r= distance between two point charge}.