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Two sphere of charge +10 and +40c are placed .12m a part.Find the position of the point between them ,when the electric field intensity is zero.Plz frnds ans correctly if u give correct ans. I will mark u as brainlist...❤ |
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Answer» Let a CHARGE of +10C placed at point A and of charge +40C paced at point B and a point between them where third charged placed be Q. And the distance between both the CHARGES is 0.12 m i.e. distance between A and B is 0.12 m. Now, assume that distance x from charge +10C. → F = (k q1 q2)/r² F1 = [k (10)(Q)]/x² .....(eq 1) F2 = [k (40)(Q)]/(0.12 - x)² ......(eq 2) → F1 = F2 → [k (10)(Q)]/x² = [k (40)(Q)]/(0.12 - x)² k and Q cancel throughout → 10/x² = 40/(0.12 - x)² → 10(0.12 - x)² = 40x² → 10(0.0144 + x² - 0.24x) = 40x² → 1(0.0144 + x² - 0.24x) = 4x² → 0.0144 + x² - 0.24x = 4x² → 3x² + 0.24x - 0.0144 = 0 → 3(x² + 0.08x - 0.0048) = 0 → x² + 0.08x - 0.0048 = 0 Using quadratic formula D = b² - 4ac → (0.08)² - 4(1)(-0.0048) → 0.64 + 4(0.0048) → 0.0064 + 0.0192 = 0.0256 x = (-b ± √D)/2a → (-0.08 ± √0.0256)/2 → (-0.08 ± 0.16)/2 → (-0.08 + 0.16)/2 [negative one is neglected] → 0.04 Therefore, x is 0.04 m AWAY from a charge Q. Coulomb's Law According to it magnitude of force between two static point charges is
i.e. F
i.e. F → F → F = (k q1 q2)/r² |
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