ElectrostaticsMedium
Question
A thin half ring of radius 35 cm is uniformly charged with a total charge of Q coulomb. If the magnitude of the electric field at centre of the half ring is $100\text{ }V/m$, then the value of Q is $\_\_\_\_$ nC.
( $\epsilon_{0} = 8.85 \times 10^{- 12}C^{2}/{Nm}^{2}$ and $\pi = 3.14$ )
Options
A.2.14
B.2.44
C.3.25
D.0.7
Solution
Sol. [Key Concept: Field at the centre of a uniformly charged half ring is $E = \frac{2k\lambda}{R}\sin\frac{\theta}{2}$ with $\theta = \pi$; solve for the total charge $Q$.]
$${E = \frac{2k\lambda}{R}sin\left( \frac{\theta}{2} \right) }{100 = 2 \times \frac{9 \times 10^{9} \times Q}{\pi R^{2}} }$$Q $= 2.14 \times 10^{- 9}C$
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