Practicing Success
Practicing Success
CUET Preparation Today
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Let electric field is zero at distance x from q1 ⇒ $\frac{kq_1}{x^2} = \frac{kq_2}{(20-x)^2}$ ⇒ $\frac{4}{x^2} = \frac{9}{(20-x)^2}$ ⇒ $\frac{2}{x} = \pm \frac{3}{20-x}$ ⇒ 40-2x = $\pm$ 3x x = 8 cm from q1
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