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Numerical · Q20

Q.A point charge of 8 μC8\ \mu\text{C} is placed in air. Find the electric potential and the electric field at a point 20 cm20\ \text{cm} from the charge.

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✓ Free question

Given q=8 μC=8×10−6 Cq=8\ \mu\text{C}=8\times 10^{-6}\ \text{C}, r=0.2 mr=0.2\ \text{m}:

V=kqr=(9×109)(8×10−6)0.2=7.2×1040.2=3.6×105 VV = \frac{kq}{r} = \frac{(9\times 10^9)(8\times 10^{-6})}{0.2} = \frac{7.2\times 10^4}{0.2} = 3.6\times 10^5\ \text{V}

E=kqr2=(9×109)(8×10−6)(0.2)2=7.2×1040.04=1.8×106 N/CE = \frac{kq}{r^2} = \frac{(9\times 10^9)(8\times 10^{-6})}{(0.2)^2} = \frac{7.2\times 10^4}{0.04} = 1.8\times 10^6\ \text{N/C}

As a check, E=V/r=(3.6×105)/0.2=1.8×106 N/CE = V/r = (3.6\times 10^5)/0.2 = 1.8\times 10^6\ \text{N/C}, which agrees.

✓Final answer

V=3.6×105 VV = 3.6\times 10^5\ \text{V}; E=1.8×106 N/CE = 1.8\times 10^6\ \text{N/C}.

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