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Question: A non-conducting ring of radius 0.5m, \(1 \cdot 11 \times {10^{ - 10}}\) coulomb charge is non-unifo...

A non-conducting ring of radius 0.5m, 1⋅11×10−101 \cdot 11 \times {10^{ - 10}} coulomb charge is non-uniformly distributed over the circumference of the ring, and produces electric field E around itself. If l=0 in the centre of the ring, then the value of ∫t=−∞0−E.dl\int\limits_{t = - \infty }^0 { - E.dl} is:
A. 2V
B. – 2V
C. – 1V
D. zero

Explanation

Solution

To answer this question, it is important to understand the concept of electrostatic potential. The electrostatic potential is defined as the work done in bringing a unit positive charge from infinity to a point in the electric field.

Complete step by step solution:
The electric field is a region of space around the charge where it exerts its sphere of influence. If any other charge comes in its vicinity, it exerts electrostatic force on the charge depending on the distance of separation from it.
The quantity electric field is defined as the electrostatic force per unit charge.

E=FqE = \dfrac{F}{q}

where q = charge in coulombs
Now, if we were to bring a unit positive charge from infinity to a specified point in the electric field, work has to be done against the electrostatic force. This work done against the electrostatic force, gets stored in the charge, represented by the quantity electric potential.
The electrical potential at a point is defined as the work done in bringing a charge from infinity to that point against the direction of the electric field.
V=WqV = \dfrac{W}{q}
Also, the potential can be calculated by the formula –
V=E×rV = E \times r
where E = electric field and r = distance of separation from the charge.
In this case, we have uniform distribution of electric field over the entire circumference of the ring. Hence, the potential can be represented by the quantity,
∫t=−∞0−E.dl=V\int\limits_{t = - \infty }^0 { - E.dl} = V
The formula for electric potential at a distance r from the charge q is given by –
V=14πε0qrV = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{r}
where ε0{\varepsilon _0}= absolute permittivity of free space.
The value of 14πε0=9×109Nm2C−2\dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}
Given,
Charge, q=1⋅1×10−10Cq = 1 \cdot 1 \times {10^{ - 10}}C
Since the potential should be calculated at the radius, the distance, r=0⋅5mr = 0 \cdot 5m
Substituting, we get –
V=9×109×1⋅1×10−100⋅5=19⋅8×10−1=1⋅98≈2VV = 9 \times {10^9} \times \dfrac{{1 \cdot 1 \times {{10}^{ - 10}}}}{{0 \cdot 5}} = 19 \cdot 8 \times {10^{ - 1}} = 1 \cdot 98 \approx 2V

Hence, electric potential at the centre of the ring is 2V.

Note: In the question, they have asked the value of a complex expression involving the integrals. Sometimes, it does not flash as to what is required in the problem. But, if you look at the options, you can see that the answers are in volts. Then, you must realise that this question is related to the concept of electric potential.