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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

  1. A 15 cm
  2. B 30 cm
  3. C 7.5 cm
  4. D 5 cm
Verified Solution

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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