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JEE Main 2023 (8 Apr, Shift 2) · Electric Potential & Equipotentials

Electric Potential & Equipotentials question from JEE Main 2023 (8 Apr, Shift 2)

Q55JEE Main 2023 · 8 APR · SHIFT 2Easy

Electric Potential & Equipotentials

Electric potential at a point 'PP' due to a point charge of 5×10−9 C5 \times 10^{-9}\text{ C} is 50 V50\text{ V}. The distance of 'PP' from the point charge is:

(Assume, 14πε0=9×10+9 N m2C−2\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{+9}\text{ N m}^2\text{C}^{-2})

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Answer: (b)

  1. V=kqr⇒r=kqVV = \frac{kq}{r} \Rightarrow r = \frac{kq}{V}
  2. r=9×109×5×10−950=4550=0.9 mr = \frac{9\times10^9 \times 5\times10^{-9}}{50} = \frac{45}{50} = 0.9\text{ m}
  3. So the answer is (b) 90 cm.
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