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Three small charges A (+2×10^-8), B(-5×10^-8) and P (+1×10^-8) lie along a line with distances AB= 6CM, BP= 4Cm and Ap=10 cm. (a) Calculate the force on the charge at P due to A and B.(b) At which point X on the line connecting A and B could there be no force on the charge P due to A and B if P were placed there?.Please no spaming |
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Answer» The Force on Charge P is 6.75 × 10⁻⁵ NThe Point X is X = 2.32 mGiven:Charge on A = 2 × 10⁻⁸ CCharge on B = - 5 × 10⁻⁸ CCharge on C = 1 × 10⁻⁸ CExplanation:As given the Distance of Separation are,⇒ AB = 6 cm = 0.06 m⇒ BP = 4 cm = 0.04 m⇒ AP = 10 cm = 0.1 MTHE Charges are kept as shown in figure:-Now, we can see that Force on P due to A will be repulsive as they are of same signs. and, Force on P due to B will be attractive as they are of opposite signs.As the Forces are ACTING at 180° angle the Resultant force will be the Subtraction of two forces,Now, lets find the Force F₁,From the FORMULA,⇒ F₁ = K (Q_{a} Q_{p}) / r²Substituting the values,⇒ F₁ = K × (2 × 10⁻⁸ × 1 × 10⁻⁸) / (0.1)²⇒ F₁ = K × (2 × 10⁻¹⁶) / (10⁻²)⇒ F₁ = K × 2 × 10⁻¹⁶ × 10²⇒ F₁ = 9 × 10⁹ × 2 × 10⁻¹⁶ × 10²⇒ F₁ = 9 × 10¹¹ × 2 × 10⁻¹⁶⇒ F₁ = 18 × 10⁽¹¹⁻¹⁶⁾⇒ F₁ = 18 × 10⁻⁵⇒ F₁ = 1.8 × 10⁻⁴⇒ F₁ = 1.8 × 10⁻⁴ NNow, Finding F₂ value,⇒ F₂ = K (Q_{b} Q_{p}) / r²Substituting the values,⇒ F₂ = K × (5 × 10⁻⁸ × 1 × 10⁻⁸) / (0.04)²⇒ F₂ = K × (5 × 10⁻¹⁶) / (4 × 10⁻²)⇒ F₂ = K × 5 × 10⁻¹⁶ × 10² / 4⇒ F₂ = 9 × 10⁹ × 5 × 10⁻¹⁶ × 10² / 4⇒ F₂ = 9 × 10¹¹ × 5 × 10⁻¹⁶ / 4⇒ F₂ = 45 × 10⁽¹¹⁻¹⁶⁾ / 4⇒ F₂ = 45 × 10⁻⁵ / 4⇒ F₂ = 11.25 × 10⁻⁵⇒ F₂ = 1.125 × 10⁻⁴⇒ F₂ = 1.125 × 10⁻⁴ NNow, let's Find the resultant force.⇒ F = F₁ - F₂(Here F Denotes Net Force)⇒ F = 1.8 × 10⁻⁴ - 1.125 × 10⁻⁴⇒ F = 10⁻⁴ (1.8 - 1.125)⇒ F = 10⁻⁴ × 0.675⇒ F = 6.75 × 10⁻⁵ ⇒ F = 6.75 × 10⁻⁵ N∴ The Force on Charge P is 6.75 × 10⁻⁵ N.Now, Point P is between A and B and we know the distance between A and B is 6 cm.Let the Point P be at X cm Distance from A and 6 - X cm Distance from BWe Know that Net Force is Zero. the force due to A is equal to force due to B.⇒ F₁ = F₂⇒ K (Q_{a} Q_{p}) / r₁² = K (Q_{b} Q_{p}) / r₂²⇒ (Q_{a} Q_{p}) / r₁² = (Q_{b} Q_{p}) / r₂²⇒ 2 × 10⁻⁸ × 1 × 10⁻⁸/(X)² = 5 × 10⁻⁸ × 1 × 10⁻⁸/(6-X)²⇒ 2 × 10⁻¹⁶ / X² = 5 × 10⁻¹⁶ / (6-X)²⇒ 2 / X² = 5 / (6 - X)²⇒ 2 × (6 - X)² = 5 × X²⇒ 2 × {36 + X² - 12 X} = 5 X²⇒ 2 X² - 24 X + 72 = 5 X²⇒ 3 X² + 24 X - 72 = 0Lets Find the Discriminant,⇒ D = b² - 4 a c⇒ D = (24)² - 4 × 3 × - 72⇒ D = 576 - 12 × - 72⇒ D = 576 + 864⇒ D = 1440.Now, Finding X Value by Quadratic Formula,⇒ X = ( - b ± √{D} ) / 2 a⇒ X = ( - 24 ± √{1440} ) / 2 × 3⇒ X = ( - 24 ± 12√{10} ) / 6⇒ X = 6 × (- 4 ± 2√{10} ) / 6⇒ X = - 4 ± 2√{10}Now,⇒ X = - 4 + 2√{10}(Here distance cannot be negative)⇒ X = - 4 + 6.32⇒ X = 2.32⇒ X = 2.32 cm∴ The Point X from A is X = 2.32 m. |
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