First slide

Electric field

Question

Charge q is uniformly distributed over a thin half ring of radius R. The electric field at the centre of the ring is

Difficult

A

$\frac{q}{2 π^{2} ε_{0} R^{2}}$
B

$\frac{q}{4 π^{2} ε_{0} R^{2}}$
C

$\frac{q}{4 {πε}_{0} R^{2}}$
D

$\frac{q}{2 {πε}_{0} R^{2}}$

Solution

$From figure, dl = Rdθ Charge on dl = λRdθ \{λ = \frac{q}{πR}\}$
Electric field at centre due to di, is $dE = k \cdot \frac{λRdθ}{R^{2}}$
We need to consider only the component dE cos $θ$ as the component dE sin $θ$ will cancel out because of the field at C due to the symmetrical element dl'.
$Total field at centre = 2 \int_{0}^{π / 2} dEcosθ$
$= \frac{2 kλ}{R} \int_{0}^{π / 2} cosθdθ = \frac{2 kλ}{R} = \frac{q}{2 π^{2} ε_{0} R^{2}}$

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