Solveeit Logo
Login

Question

Question: A non-conducting ring of radius 0.5m carries a total charge of \(1.11\times { 10 }^{ -10 }C\), distr...

A non-conducting ring of radius 0.5m carries a total charge of 1.11×10−10C1.11\times { 10 }^{ -10 }C, distributed non-uniformly on its circumference, producing an electric field in space. What is the value of the line integral −∫∞1E→.dl→-\int _{ \infty }^{ 1 }{ \overrightarrow { E } .\overrightarrow { dl } } (l=0 being the centre of the ring), in volts.

Explanation

Solution

Use the definition of potential difference. The substitute potential at infinity is zero. Then use the potential formula. Substitute the values in the expression and calculate the value of the integral.

Complete step-by-step solution:
Given: Total charge Q=1.11×10−10C1.11\times { 10 }^{ -10 }C
Radius of ring R=0.5m
By the definition of potential difference we know,
−∫∞1E→.dl→=∫∞1dV-\int _{ \infty }^{ 1 }{ \overrightarrow { E } .\overrightarrow { dl } } = \int _{ \infty }^{ 1 }{ dV }
∴−∫∞1E→.dl→=∫∞1dV\therefore -\int _{ \infty }^{ 1 }{ \overrightarrow { E } .\overrightarrow { dl } } =\int _{ \infty }^{ 1 }{ dV }
∴−∫∞1E→.dl→=Vl=0−Vl=∞\therefore -\int _{ \infty }^{ 1 }{ \overrightarrow { E } .\overrightarrow { dl } } =\quad { V }_{ l=0 }-{ V }_{ l=\infty }
But we know, potential at infinity is zero.
∴−∫∞1E→.dl→=Vl=0\therefore -\int _{ \infty }^{ 1 }{ \overrightarrow { E } .\overrightarrow { dl } } =\quad { V }_{ l=0 } …………………..(1)
The potential at the center is given by,
Vl=0=14πϵ0QR{ V }_{ l=0 }=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { Q }{ R }...........(2)
But, from equation.(1) we know,
−∫∞1E→.dl→=Vl=0-\int _{ \infty }^{ 1 }{ \overrightarrow { E } .\overrightarrow { dl } } =\quad { V }_{ l=0 }
Therefore, equation.(2) becomes,
−∫∞1E→.dl→=14πϵ0QR-\int _{ \infty }^{ 1 }{ \overrightarrow { E } .\overrightarrow { dl } } =\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \cfrac { Q }{ R }
Now, by substituting values in above equation, we get,
−∫∞1E→.dl→=14πϵ01.1×10−100.5-\int _{ \infty }^{ 1 }{ \overrightarrow { E } .\overrightarrow { dl } } =\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { 1.1\times { 10 }^{ -10 } }{ 0.5 }
∴−∫∞1E→.dl→=9×109×1.1×10−100.5\therefore -\int _{ \infty }^{ 1 }{ \overrightarrow { E } .\overrightarrow { dl } } =9\times { 10 }^{ 9 }\times \dfrac { 1.1\times { 10 }^{ -10 } }{ 0.5 }
∴−∫∞1E→.dl→=1.98\therefore -\int _{ \infty }^{ 1 }{ \overrightarrow { E } .\overrightarrow { dl } } =1.98
Hence, the value of line integral −∫∞1E→.dl→-\int _{ \infty }^{ 1 }{ \overrightarrow { E } .\overrightarrow { dl } } 1s 1.98.

Note: Potential at the center of the ring remains the same irrespective of whether the charge distribution is uniform or non-uniform. It is a common convention to set the electric potential of a charge at infinity to be zero. The reason behind this convention can be that the electric potential is inversely proportional to distance. So, as the distance increases, potential decreases.