Electrostatics

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53 of 311

Question 53 of 311

A thin half ring of radius 3535 cm is uniformly charged with a total charge of QQ coulomb. If the magnitude of the electric field at centre of the half ring is 100100 V/m, then the value of QQ is _______ nC. (ϵo=8.85×10−12\epsilon_o = 8.85 \times 10^{-12} C2^2/Nm2^2 and π=3.14\pi = 3.14)

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Answer: (A) 2.142.14

The electric field at the center of a uniformly charged half ring is given by:

E=2kλRE = \dfrac{2k\lambda}{R}

where λ=QπR\lambda = \dfrac{Q}{\pi R} is the linear charge density.

Substituting λ\lambda, we get:

E=2kQπR2=2Q4πϵ0πR2=Q2π2ϵ0R2E = \dfrac{2kQ}{\pi R^2} = \dfrac{2Q}{4\pi\epsilon_0 \pi R^2} = \dfrac{Q}{2\pi^2\epsilon_0 R^2}

Rearranging for QQ:

Q=2π2ϵ0R2EQ = 2\pi^2\epsilon_0 R^2 E

Given values:
E=100E = 100 V/m
R=35R = 35 cm =0.35= 0.35 m
ϵ0=8.85×10−12\epsilon_0 = 8.85 \times 10^{-12} C2^2/Nm2^2
π=3.14\pi = 3.14

Substituting the values:

Q=2×(3.14)2×8.85×10−12×(0.35)2×100Q = 2 \times (3.14)^2 \times 8.85 \times 10^{-12} \times (0.35)^2 \times 100

Q=2×9.8596×8.85×10−12×0.1225×100Q = 2 \times 9.8596 \times 8.85 \times 10^{-12} \times 0.1225 \times 100

Q≈2137.8×10−12Q \approx 2137.8 \times 10^{-12} C

Q≈2.14×10−9Q \approx 2.14 \times 10^{-9} C =2.14= 2.14 nC

Answer: 2.142.14

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