Q.40. Two point charges and are located at and , respectively on the x-axis. The ratio of electri

Q: What is the correct answer for this exam Physics Q40?

The correct answer is C. According to Gauss's Law, the electric flux through a closed surface is $\Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0}$. For the first sphere: Radius $R_1 = 3 \mathrm{~cm}$. Its center is at the origin. The charges are $q_1 = 8 \mu \mathrm{C}$ at $x = 2 \mathrm{~cm}$ and $q_2 = -2 \mu \mathrm{C}$ at $x = 4 \mathrm{~cm}$. For $R_1 = 3 \mathrm{~cm}$, only $q_1$ is enclosed because $2 \mathrm{~cm} <

Q: Q.40. Two point charges and are located at and , respectively on the x-axis. The ratio of electric flux due to these charges through two spheres of radii and with their centers at the origin is…

The correct answer is C. According to Gauss's Law, the electric flux through a closed surface is $\Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0}$. For the first sphere: Radius $R_1 = 3 \mathrm{~cm}$. Its center is at the origin. The charges are $q_1 = 8 \mu \mathrm{C}$ at $x = 2 \mathrm{~cm}$ and $q_2 = -2 \mu \mathrm{C}$ at $x = 4 \mathrm{~cm}$. For $R_1 = 3 \mathrm{~cm}$, only $q_1$ is enclosed because $2 \mathrm{~cm} <

Question

Q.40. Two point charges  and  are located at  and , respectively on the x-axis. The ratio of electric flux due to these charges through two spheres of radii  and  with their centers at the origin is __________

FirstInTest · Accepted Answer

According to Gauss's Law, the electric flux through a closed surface is $\Phi = \frac{Q_{	ext{enclosed}}}{\epsilon_0}$.
For the first sphere: Radius $R_1 = 3 \mathrm{~cm}$. Its center is at the origin.
The charges are $q_1 = 8 \mu \mathrm{C}$ at $x = 2 \mathrm{~cm}$ and $q_2 = -2 \mu \mathrm{C}$ at $x = 4 \mathrm{~cm}$.
For $R_1 = 3 \mathrm{~cm}$, only $q_1$ is enclosed because $2 \mathrm{~cm} < 3 \mathrm{~cm}$ but $4 \mathrm{~cm} > 3 \mathrm{~cm}$.
So, $Q_{	ext{enclosed,1}} = q_1 = 8 \mu \mathrm{C}$.
$\Phi_1 = \frac{8 \mu \mathrm{C}}{\epsilon_0}$.
For the second sphere: Radius $R_2 = 5 \mathrm{~cm}$. Its center is at the origin.
For $R_2 = 5 \mathrm{~cm}$, both $q_1$ and $q_2$ are enclosed because $2 \mathrm{~cm} < 5 \mathrm{~cm}$ and $4 \mathrm{~cm} < 5 \mathrm{~cm}$.
So, $Q_{	ext{enclosed,2}} = q_1 + q_2 = 8 \mu \mathrm{C} - 2 \mu \mathrm{C} = 6 \mu \mathrm{C}$.
$\Phi_2 = \frac{6 \mu \mathrm{C}}{\epsilon_0}$.
The ratio $\frac{\Phi_1}{\Phi_2} = \frac{8 \mu \mathrm{C} / \epsilon_0}{6 \mu \mathrm{C} / \epsilon_0} = \frac{8}{6} = \frac{4}{3}$.
The ratio is $4:3$.

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