JEE MAIN - Physics (2025 - 29th January Evening Shift - No. 17)
A point charge causes an electric flux of $-2 \times 10^4 \mathrm{Nm}^2 \mathrm{C}^{-1}$ to pass through a spherical Gaussian surface of 8.0 cm radius, centred on the charge. The value of the point charge is :
(Given $\epsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$ )
Explanation
Given an electric flux $\phi = -2 \times 10^4 \, \text{Nm}^2\text{C}^{-1}$ 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 $Q$. According to Gauss's Law:
$ \phi = \frac{Q}{\varepsilon_0} $
Where $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2\text{N}^{-1}\text{m}^{-2}$.
Rearranging the equation to solve for $Q$, we have:
$ Q = \phi \varepsilon_0 $
Substituting the given values:
$ Q = -2 \times 10^4 \times 8.85 \times 10^{-12} $
Calculating this yields:
$ Q = -17.7 \times 10^{-8} \, \text{C} $
Thus, the value of the point charge is $Q = -17.7 \times 10^{-8} \, \text{C}$.
Comments (0)
