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A non conducting ring of radius 0.5 \mathrm{~m} carries a total charge of 1.11 \times 10^{-10} \mathrm{C} distributed non-aniformly on its circumference producing an electric field \mathrm{E} everywhere in space. The value of the integral \int_{l=\infty}^{l=0}-\vec{E} \cdot \overrightarrow{d l}(l=0 \: \text{being, centre of the ring}) is volt is

Option: 1

+2


Option: 2

5


Option: 3

8


Option: 4

Zero


Answers (1)

best_answer

Use  \quad-\int_{l=\infty}^{l=0} \vec{E} \cdot d \vec{l}=\int_{l=\infty}^{l=0} d v=v(c)-v(I)
                                                           \uparrow                    \uparrow
                                                          centre              infinity

but v(infinity)= 0

\therefore \quad-\int_{l=\infty}^{l=0} \vec{E} \cdot d \vec{l}  corresponds to potential at centre of ring and v\left ( \text{centre} \right )= \frac{1}{4 \pi \epsilon _{0}}\cdot \frac{q}{R}
                                                                                                                  v(c)=\frac{9 \times 10^{9} \times 1.11 \times 10^{-10}}{0.5}
v(c)=2 \text { volt }  Ans 

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