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Electrostatics question

2023 · 8 Apr · Shift 2 · Q52
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  5. /2023 · 8 Apr · Shift 2 · Q52

Electrostatics question

2023 · 8 Apr · Shift 2 · Q52

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Electric potential at a point 'P\mathrm{P}P' due to a point charge of 5×10−9C5 \times 10^{-9} \mathrm{C}5×10−9C is 50 V50 \mathrm{~V}50 V. The distance of 'P\mathrm{P}P' from the point charge is: (Assume, 14πε0=9×10+9 Nm2C−2\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{+9} ~\mathrm{Nm}^{2} \mathrm{C}^{-2}4πε0​1​=9×10+9 Nm2C−2 )
  1. A
    0.9 cm
  2. B
    90 cm
  3. C
    3 cm
  4. D
    9 cm
View written solutionFree

Correct answer: B

  1. Use the formula for electric potential due to a point charge

    V=14πε0⋅qrV = \frac{1}{4\pi\varepsilon_0}\cdot \frac{q}{r}V=4πε0​1​⋅rq​

    Given: V=50 V,q=5×10−9 C,14πε0=9×109V = 50\,\text{V}, \quad q = 5\times 10^{-9}\,\text{C}, \quad \frac{1}{4\pi\varepsilon_0} = 9\times 10^9V=50V,q=5×10−9C,4πε0​1​=9×109

  2. Substitute the values

    50=9×109×5×10−9r50 = \frac{9\times 10^9 \times 5\times 10^{-9}}{r}50=r9×109×5×10−9​

  3. Simplify the numerator

    9×109×5×10−9=459\times 10^9 \times 5\times 10^{-9} = 459×109×5×10−9=45

    So, 50=45r50 = \frac{45}{r}50=r45​

  4. Solve for rrr

    r=4550=0.9 mr = \frac{45}{50} = 0.9\,\text{m}r=5045​=0.9m

  5. Convert to cm

    0.9 m=90 cm0.9\,\text{m} = 90\,\text{cm}0.9m=90cm

  6. Match with the options

    90 cm\boxed{90\,\text{cm}}90cm​

    Hence, the correct option is B.

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