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JEE Main 2019 (11 Jan, Shift 1) · Electrostatic Potential Energy

Electrostatic Potential Energy question from JEE Main 2019 (11 Jan, Shift 1)

Q50JEE Main 2019 · 11 JAN · SHIFT 1Medium

Electrostatic Potential Energy

Three charges Q, + q and + q are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is:

Figure

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Answer: (d)

  1. Let each leg be aa, so the hypotenuse is 2a\sqrt{2}a. QQ is aa from one +q+q and 2a\sqrt{2}a from the other; the two +q+q are aa apart.
  2. U=k[qQa+qQ2a+q2a]=0U = k\left[\frac{qQ}{a} + \frac{qQ}{\sqrt{2}a} + \frac{q^2}{a}\right] = 0
  3. Q(1+12)=−q⇒Q=−2q2+1Q\left(1 + \frac{1}{\sqrt{2}}\right) = -q \Rightarrow Q = -\frac{\sqrt{2}q}{\sqrt{2}+1}
  4. So the answer is (d).
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