SATHEE: Electric Charges and Fields - Result Question 26

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 $λ$ . The electric field at the centre of this arc is

[2000]

(a) $\frac{λ}{2 π ε_{0} a}$

(b) $\frac{λ}{2 π ε_{0} a^{2}}$

(c) $\frac{λ}{4 π^{2} ε_{0} a}$

(d) $\frac{λ^{2}}{2 π ε_{0} a}$

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Answer:

Correct Answer: 26. (a)

Solution:

(a) $λ =$ linear charge density;

Charge on elementary portion $d x = λ d x$ .

Electric field at $O, d E = \frac{λ d x}{4 π ε_{0} a^{2}}$

Horizontal electric field, i.e., perpendicular to $A O$ , will be cancelled.

Hence, net electric field $=$ addition of all electrical fields in direction of $A O$

$= Σ d E \cos θ$

$\Rightarrow E = \int \frac{λ d x}{4 π ε_{0} a^{2}} \cos θ$

Also, $d θ = \frac{d x}{a}$ or $d x = a d θ$

$E = \int_{- π / 2}^{π / 2} \frac{λ \cos θ d θ}{4 π ε_{0} a} = \frac{λ}{4 π ε_{0} a} [\sin θ]_{- π / 2}^{π / 2}$

$= \frac{λ}{4 π ε_{0} a} [1 - (- 1)] = \frac{λ}{2 π ε_{0} a}$

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