Two identical charged particles each having a mass 10g and charge 2.0 × 10^-7 C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10ms^-2]

Correct Option - 1


Detailed Solution:

$\mu mg=\frac{k{q}_1{q}_2}{L^2}$

$\Rightarrow 0.25\times 0.01\times 10=\frac{9\times 10^9\times 2\times 10^{-7}\times 2\times 10^{-7}}{L^2}$

$=\frac{9\times 2^2\times 10^{-5+4}}{25\times 10}$

L = 12 cm

<p><math><mrow><mi>μ</mi><mi>m</mi><mi>g</mi><mo>=</mo><mfrac><mrow><mi>k</mi><msub><mrow><mi>q</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>q</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow><mrow><msup><mrow><mi>L</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></math></p><p><math><mrow><mo>⇒</mo><mn>0</mn><mo>.</mo><mn>2</mn><mn>5</mn><mo>×</mo><mn>0</mn><mo>.</mo><mn>0</mn><mn>1</mn><mo>×</mo><mn>1</mn><mn>0</mn><mo>=</mo><mfrac><mrow><mn>9</mn><mo>×</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>9</mn></mrow></msup><mo>×</mo><mn>2</mn><mo>×</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mo>−</mo><mn>7</mn></mrow></msup><mo>×</mo><mn>2</mn><mo>×</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mo>−</mo><mn>7</mn></mrow></msup></mrow><mrow><msup><mrow><mi>L</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></math></p><p><math><mrow><mo>=</mo><mfrac><mrow><mn>9</mn><mo>×</mo><msup><mrow><mn>2</mn></mrow><mrow><mn>2</mn></mrow></msup><mo>×</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mo>−</mo><mn>5</mn><mo>+</mo><mn>4</mn></mrow></msup></mrow><mrow><mn>2</mn><mn>5</mn><mo>×</mo><mn>1</mn><mn>0</mn></mrow></mfrac></mrow></math></p><p>L = 12 cm</p><div class="photo-widget-full"><div class="figure"><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1753177332phpzVlvke.jpeg" alt="" width="191" height="150" loading="lazy"></picture></div></div>

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