two charges 20×10^-6C and 1.0×10^-6 are placed at a sepration of 10 cm . where should a third charge be placed such that it experiences no net force due to these charges?
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Let third charge is placed x m from first charge .
According to question,
No net force due to charges , means Force acts on 3rd due to 1st = Force acts on 3rd due to 2nd
so, KQ₁Q₃/x² = KQ₂Q₃/(10 - x)²
Q₁/x² = Q₂/(10 - x)²
Taking square root both sides,
√Q₁/x = √Q₂/(10 - x)
x = √Q₁/(√Q₁ + √Q₂) × 10
Put Q₁ = 2.0 × 10⁻⁶C and Q₂ = 1.0 × 10⁻⁶C
x = √2/(√2 + 1) × 10 = 14.4/(2.44) cm = 5.9cm
Hence, 3rd charge is placed 5.9 cm from 1st charge
According to question,
No net force due to charges , means Force acts on 3rd due to 1st = Force acts on 3rd due to 2nd
so, KQ₁Q₃/x² = KQ₂Q₃/(10 - x)²
Q₁/x² = Q₂/(10 - x)²
Taking square root both sides,
√Q₁/x = √Q₂/(10 - x)
x = √Q₁/(√Q₁ + √Q₂) × 10
Put Q₁ = 2.0 × 10⁻⁶C and Q₂ = 1.0 × 10⁻⁶C
x = √2/(√2 + 1) × 10 = 14.4/(2.44) cm = 5.9cm
Hence, 3rd charge is placed 5.9 cm from 1st charge
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