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Question: A semi-circular arc of radius ‘a’ is charged uniformly and the charge per unit length is \(\lambda\)...

A semi-circular arc of radius ‘a’ is charged uniformly and the charge per unit length is λ\lambda. The electric field at the centre of this arc is​
a. λ2πε0a\dfrac{\lambda }{{2\pi {\varepsilon _0}a}}
b. λ2πε0a2\dfrac{\lambda }{{2\pi {\varepsilon _0}{a^2}}}
c. λ4π2ε0a\dfrac{\lambda }{{4{\pi ^2}{\varepsilon _0}a}}
d. λ22πε0a\dfrac{{{\lambda ^2}}}{{2\pi {\varepsilon _0}a}}

Explanation

Solution

The charge is the property of particles which attracts and repulses with each other when placed in an electrical field. Rate of flow of charges is called current. Electrification is the process of adding charge to the body. Apply the electric field at the center of the arc.

Complete step by step answer:
Electrification is the process of adding charge to the body. The property of the matter that produces and experiences electrical and magnetic fields is called charge.
Voltage is the product of current and resistance. The charge is the property of particles which attract and repulse with each other when placed in an electrical field.
A charged particle exerts force on each other. A 1.6×10191.6 \times {10^{19}} is the charge of a single electron. Electronic circuits use electric charge to do some useful work.
The moving charges produce an electric current. Rate of flow of charges is called current. The SI unit of current is ampere. One ampere equals one coulomb per second.
Charge on the elementary portion dx=λdxdx = \lambda dx
Where λ\lambda is the charge density.

Then the electric field is given by
⇒dE=λdx4πε0a2\Rightarrow dE = \dfrac{{\lambda dx}}{{4\pi {\varepsilon _0}{a^2}}}
Here, horizontal electric is cancelled since it is perpendicular.
Hence, the net electric field is equal to the addition of all electrical fields
⇒E=∫λdx4πε0a2cos⁡θ\Rightarrow E = \int {\dfrac{{\lambda dx}}{{4\pi {\varepsilon _0}{a^2}}}} \cos \theta
Integrating we get, E=∫−π2π2λcos⁡θdθ4πε0a=λ4πε0a[1−(−1)]=λ2πε0aE = \int\limits_{\frac{{-{\pi}}}{2}}^{\frac{\pi }{2}} {\dfrac{{\lambda \cos \theta d\theta }}{{4\pi {\varepsilon _0}a}}} = \dfrac{\lambda }{{4\pi {\varepsilon _0}a}}\left[ {1 - \left( { - 1} \right)} \right] = \dfrac{\lambda }{{2\pi {\varepsilon _0}a}}

Hence, the correct answer is option (A).

Note: Fundamental property that attracts or repulses each other when placed in a magnetic field is called charge. When the linear size of a charged body is smaller than the distance, then the size can be ignored and that charge body is called point charge. Charge experiences the field only in the electric field.