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The electric field strength due to a point charge of 5 μC at a distance of 80 cm from the charge is

1. 4 × 10^8 N/C
2. 4 × 10^7 N/C
3. 4 × 10^5 N/C
4. 4 × 10^4 N/C

Correct answer: 4 × 10^5 N/C

Solution

For a point charge, \(E = kQ/r^2\). Here \(Q = 5\times10^{-6}\,\text{C}\) and \(r = 0.8\,\text{m}\). Substituting gives \(E = 9\times10^9 \times 5\times10^{-6} / 0.64 \approx 7.0\times10^4\,\text{N/C}\), so the OCR/source options appear inconsistent; the marked option is \(4\times10^7\,\text{N/C}\), but the physically correct value from the stated data is about \(7\times10^4\,\text{N/C}\).

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