TS EAMCET · Physics · Electrostatics
Two point charges of magnitudes \(-8 \mu \mathrm{C}\) and \(+32 \mu \mathrm{C}\) are separated by a distance of 15 cm in air. The position of the point from \(-8 \mu \mathrm{C}\) charge at which the resultant electric field becomes zero is
- A 15 cm
- B 30 cm
- C 7.5 cm
- D 5 cm
Answer & Solution
Correct Answer
(A) 15 cm
Step-by-step Solution
Detailed explanation
Let the charge be placed at a distance \(x\) from \(-8 \mu \mathrm{c}\) at which resultant electric field becomes zero \(\begin{aligned} & \therefore \frac{8}{4 \pi \varepsilon_0 x^2}=\frac{32}{4 \pi \varepsilon_0(15+x)^2} \\ & \Rightarrow \frac{x+15}{x}=2 \end{aligned}\) On…
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