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QNEET-2024-Ph-145: NEET 2024 Physics - Class 12 NEET-2024-Ph-145 Solution: The value of electric potential at a distance of 9 cm from the…

The value of electric potential at a distance of 9 cm from the point charge 4×10−7 C4 \times 10^{-7}\ \text{C} is [Given 14πϵ0=9×109 N m2 C−2\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\ \text{N m}^2\ \text{C}^{-2}]: 4×102 V4 \times 10^2\ \text{V} 44.4 V 4.4×105 V4.4 \times 10^5\ \text{V} 4×104 V4 \times 10^4\ \text{V}

Multiple Choice Options

  • A 4×102 V4 \times 10^2 \mathrm{~V} 4 _ 1 0 2   V
  • B 44.4 V44.4 \mathrm{~V} 44.4   V
  • C 4.4×105 V4.4 \times 10^5 \mathrm{~V} 4.4 _ 1 0 5   V
  • D 4×104 V4 \times 10^4 \mathrm{~V} 4 _ 1 0 4   V

Step-by-Step Verified Solution

The electric potential (V) at a distance (r) from a point charge (q) is given by: V = 1 4 � _ 0 q r where 1 4 � _ 0 is Coulomb's constant. In this case, we have: q = 4 _ 10 _ 7 C r = 9   cm = 0.09   m 1 4 � _ 0 = 9 _ 10 9   N   m 2 C _ 2 Substituting these values into the equation for electric potential, we get: V = ( 9 _ 10 9   N   m 2 C _ 2 ) ( 4 _ 10 _ 7 C ) ( 0.09   m ) V = 4 _ 10 4   V Therefore, the value of electric potential at a distance of 9   cm from the point charge 4 _ 10 _ 7 C is 4 _ 10 4   V . The correct answer is Option D.

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