Verified Solution • 2025–26 Session moderate

QJEEM25Pheb9a: JEE Main 2025 Physics - Class 12 JEEM25Pheb9a Solution: A point charge causes an electric flux of to pass through a…

A point charge causes an electric flux of −2×104Nm2C−1-2 \times 10^4 \mathrm{Nm}^2 \mathrm{C}^{-1} to pass through a spherical Gaussian surface of 8.0 cm radius, centred on the charge. The value of the point charge is :

(Given ϵ0=8.85×10−12C2 N−1 m−2\epsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2} )

Multiple Choice Options

  • A 17.7×10−8C17.7 \times 10^{-8} \mathrm{C}
  • B −17.7×10−8C-17.7 \times 10^{-8} \mathrm{C}
  • C 15.7×10−8C15.7 \times 10^{-8} \mathrm{C}
  • D −15.7×10−8C-15.7 \times 10^{-8} \mathrm{C}

Step-by-Step Verified Solution

Given an electric flux ϕ=−2×104 Nm2C−1\phi = -2 \times 10^4 \, \text{Nm}^2\text{C}^{-1} through a spherical Gaussian surface with a radius of 8.0 cm (r = 0.08 m), we are to find the value of the point charge QQ . According to Gauss's Law:

ϕ=Qε0\phi = \frac{Q}{\varepsilon_0}

Where ε0=8.85×10−12 C2N−1m−2\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2\text{N}^{-1}\text{m}^{-2} .

Rearranging the equation to solve for QQ , we have:

Q=ϕε0Q = \phi \varepsilon_0

Substituting the given values:

Q=−2×104×8.85×10−12Q = -2 \times 10^4 \times 8.85 \times 10^{-12}

Calculating this yields:

Q=−17.7×10−8 CQ = -17.7 \times 10^{-8} \, \text{C}

Thus, the value of the point charge is Q=−17.7×10−8 CQ = -17.7 \times 10^{-8} \, \text{C} .

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