Question

Two point charges of +3uc each are 100 cm apart . At what point on the line joining the charges will the electric intensity be zero

Answer 1

Answer:

50cm

Explanation:

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Answer 2

Answer:

$\bf{given \: charges \: \implies \: 3 </p><p> \mu \: c \: = 3 \times {10}^{ - 6} } \\ \\ \bf{ distance \: between \: these \: charges \: = 100cm} \\ \\ let \: \\ \bf{ at \: x \: cm \: \: electric \: field \: intensity \: be \: zero} \\ \\ \bf{and \: this \: point \: is \: between \: these \: charges} \\ \\ \therefore \bf{\: its \: distance \: is \:( 100 - x \: ) \: cm} \\ \\ \bf{ if \: electric \: field \: be \: equal} \\ \\ \implies \frac{k \times 3 \times {10}^{ - 6} }{ {x}^{2} } = \frac{k \times 3 \times {10}^{ - 6} }{ {(100 - x)}^{2} } \\ \\ \implies \: \frac{1}{ {x}^{2} } = \frac{1}{ {(100 - x)}^{2} } \\ \\ \because{100cm = 1m} \\ \\ \implies \: 2x = 1 \\ \\ \implies \: x = \frac{1}{2} \: \: metre \: or \: \frac{100}{2} = 50 \: cm.$