Electric Charges and Fields - Result Question 26
26. A semi-circular arc of radius ’ $a$ ’ is charged uniformly and the charge per unit length is $\lambda$. The electric field at the centre of this arc is
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(a) $\frac{\lambda}{2 \pi \varepsilon_0 a}$
(b) $\frac{\lambda}{2 \pi \varepsilon_0 a^{2}}$
(c) $\frac{\lambda}{4 \pi^{2} \varepsilon_0 a}$
(d) $\frac{\lambda^{2}}{2 \pi \varepsilon_0 a}$
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Answer:
Correct Answer: 26. (a)
Solution:
- (a) $\lambda=$ linear charge density;
Charge on elementary portion $d x=\lambda d x$.
Electric field at $O, d E=\frac{\lambda d x}{4 \pi \varepsilon_0 a^{2}}$
Horizontal electric field, i.e., perpendicular to $A O$, will be cancelled.
Hence, net electric field $=$ addition of all electrical fields in direction of $A O$
$=\Sigma d E \cos \theta$
$\Rightarrow E=\int \frac{\lambda d x}{4 \pi \varepsilon_0 a^{2}} \cos \theta$
Also, $d \theta=\frac{d x}{a}$ or $d x=a d \theta$
$E=\int _{-\pi / 2}^{\pi / 2} \frac{\lambda \cos \theta d \theta}{4 \pi \varepsilon_0 a}=\frac{\lambda}{4 \pi \varepsilon_0 a}[\sin \theta] _{-\pi / 2}^{\pi / 2}$
$=\frac{\lambda}{4 \pi \varepsilon_0 a}[1-(-1)]=\frac{\lambda}{2 \pi \varepsilon_0 a}$
