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Q.A sphere of radius 5 cm has a charge of 31.41 μC. Calculate the surface density of charge.

Bihar BsebBihar Board Intermediate 2023Subjective· 2mImportance★★★★★
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Surface density = charge ÷ surface area of the sphere. With Q=31.41 μCQ=31.41\,\mu C and r=5r=5 cm, σ≈1.0×10−3\sigma \approx 1.0\times10^{-3} C/m².

Surface charge density is the charge per unit surface area:

σ=QA=Q4πr2.\sigma = \frac{Q}{A} = \frac{Q}{4\pi r^2}.

Given: Q=31.41 μC=31.41×10−6Q = 31.41\ \mu C = 31.41\times10^{-6} C, r=5r = 5 cm =0.05= 0.05 m.

Surface area:

A=4πr2=4π(0.05)2=4π(2.5×10−3)=3.1416×10−2 m2.A = 4\pi r^2 = 4\pi (0.05)^2 = 4\pi (2.5\times10^{-3}) = 3.1416\times10^{-2}\ \text{m}^2.

Surface density: …

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