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5) Two charges + 2 kee and the are separated by adistance 10cm apart in air where should a thirdcharge tshic be placed so that the charge is inequilibriumAlso and intensity of electric field at a point medwayb/w the charges. |
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Answer» Answer: Given conditions ⇒ q₁ = 12 μC = 12 × 10⁻⁶ C. q₂ = 8 μC = 8 × 10⁻⁶ C. Distance between them(r₁) = 10 cm. = 0.1 m. Final Distance between them(r₂) = 6 cm . = 0.06 m. Work done in bringing the charges upto the distance of 6 cm is given by, W = k q₁ q₂ (1/r₂ - 1/r₁) = 9 × 10⁹ × 12 × 10⁻⁶ × 8 × 10⁻⁶(1/0.06 - 1/0.1) = 864 × 10⁻³ (16.67 - 10) = 864 × 10⁻³(6.67) = 5760 × 10⁻³ = 5.76 J. |
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