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A half ring of radius R has a charge of per unit length. The electric field at the centre is \left ( k=\frac{1}{4\pi \varepsilon _{0}} \right )  

  • Option 1)

    Zero

  • Option 2)

    \frac{k\lambda }{R}

  • Option 3)

    \frac{2k\lambda }{R}

  • Option 4)

    \frac{k\pi \lambda }{R}

 

Answers (1)

As we learned

 

Charged Circular ring -

A charged circular ring of radius R and charge Q.

- wherein

 

 Charge on dl = \lambda Rd\theta .

Field at C due to dl = k\frac{\lambda Rd\theta }{R^{2}}=dE

We need to consider only the component dE\cos \theta ,  as the component dE \sin \theta will cancel out because of the field at C due to the symmetrical element dl¢,

The total field at C is  =2\int_{0}^{\frac{\pi }{2}}  dE \cos \theta =2\frac{k\lambda }{R}\int_{0}^{\frac{\pi }{2}} \cos \theta d\theta =2k\frac{\lambda }{R}


Option 1)

Zero

Option 2)

\frac{k\lambda }{R}

Option 3)

\frac{2k\lambda }{R}

Option 4)

\frac{k\pi \lambda }{R}

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Vakul

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