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GUJCET · Physics · Electric Charges and Fields
Two point unlike charges having magnitude \(+16 \mu C\) and \(-9 \mu C\) are separated by a distance 10 cm in air. The resultant electric field will be zero at distance _________ from \(-9 \mu C\) charge.
- A 40 cm
- B 20 cm
- C 10 cm
- D 30 cm
Answer & Solution
Correct Answer
(D) 30 cm
Step-by-step Solution
Detailed explanation
(D)
\(q_1=16 \mu C , q_2=-9 \mu C\)
Let \(x\) be the position from \(-9 \mu C\) charge where resultant electric field is zero.
\(
\therefore E_1=E_2
\)
\(\begin{array}{l}\therefore \frac{k q_1}{(10+x)^2 \times 10^{-4}}=\frac{k q_2}{x^2 \times 10^{-4}} \\ \therefore \frac{k \times 16 \times 10^{-6}}{(10+x)^2 \times 10^{-4}}=\frac{k \times 9 \times 10^{-6}}{x^2 \times 10^{-4}}\end{array}\)
\(\begin{array}{l}\therefore \frac{4}{10+x}=\frac{3}{x} \\ \therefore 4 x=30+3 x \\ \therefore x=30 cm\end{array}\)
\(q_1=16 \mu C , q_2=-9 \mu C\)
Let \(x\) be the position from \(-9 \mu C\) charge where resultant electric field is zero.
\(
\therefore E_1=E_2
\)
\(\begin{array}{l}\therefore \frac{k q_1}{(10+x)^2 \times 10^{-4}}=\frac{k q_2}{x^2 \times 10^{-4}} \\ \therefore \frac{k \times 16 \times 10^{-6}}{(10+x)^2 \times 10^{-4}}=\frac{k \times 9 \times 10^{-6}}{x^2 \times 10^{-4}}\end{array}\)
\(\begin{array}{l}\therefore \frac{4}{10+x}=\frac{3}{x} \\ \therefore 4 x=30+3 x \\ \therefore x=30 cm\end{array}\)
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