Question:
A large number of positive charges each of magnitude \(q\) are placed along the \(x\)-axis at the origin and at every \(1\,\text{cm}\) distance in both the directions. The electric flux through a spherical surface of radius \(2.5\,\text{cm}\) centered at the origin is
A large number of positive charges each of magnitude \(q\) are placed along the \(x\)-axis at the origin and at every \(1\,\text{cm}\) distance in both the directions. The electric flux through a spherical surface of radius \(2.5\,\text{cm}\) centered at the origin is
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According to Gauss's law, electric flux through a closed surface depends only on the total enclosed charge and not on the positions of the charges inside the surface.
Updated On: Jun 22, 2026
- \(\dfrac{5q}{\varepsilon_0}\)
- \(\dfrac{8q}{\varepsilon_0}\)
- \(0\)
- \(\infty\)
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The Correct Option is A
Solution and Explanation
Step 1: Use Gauss's law.
According to Gauss's law, the total electric flux through a closed surface is \[ \Phi=\frac{Q_{\text{enclosed}}}{\varepsilon_0} \] Therefore, we need to determine the total charge enclosed inside the sphere.
Step 2: Identify the charges inside the sphere.
The sphere has radius \[ 2.5\,\text{cm} \] Charges are placed on the \(x\)-axis at intervals of \(1\,\text{cm}\) in both directions including the origin.
Thus, the charge positions are: \[ x=0,\pm1,\pm2,\pm3,\ldots \] Only those charges whose distance from the origin is less than or equal to \(2.5\,\text{cm}\) lie inside the sphere.
Hence, the enclosed charges are at: \[ x=0,\pm1,\pm2 \] The charges at \[ x=\pm3 \] lie outside the sphere.
Therefore, total number of enclosed charges is \[ 5 \]
Step 3: Calculate the enclosed charge.
Each charge has magnitude \[ q \] Hence, \[ Q_{\text{enclosed}}=5q \]
Step 4: Calculate the electric flux.
Using Gauss's law, \[ \Phi=\frac{Q_{\text{enclosed}}}{\varepsilon_0} \] \[ \Phi=\frac{5q}{\varepsilon_0} \]
Step 5: Final conclusion.
Hence, the electric flux through the spherical surface is \[ \boxed{\frac{5q}{\varepsilon_0}} \]
According to Gauss's law, the total electric flux through a closed surface is \[ \Phi=\frac{Q_{\text{enclosed}}}{\varepsilon_0} \] Therefore, we need to determine the total charge enclosed inside the sphere.
Step 2: Identify the charges inside the sphere.
The sphere has radius \[ 2.5\,\text{cm} \] Charges are placed on the \(x\)-axis at intervals of \(1\,\text{cm}\) in both directions including the origin.
Thus, the charge positions are: \[ x=0,\pm1,\pm2,\pm3,\ldots \] Only those charges whose distance from the origin is less than or equal to \(2.5\,\text{cm}\) lie inside the sphere.
Hence, the enclosed charges are at: \[ x=0,\pm1,\pm2 \] The charges at \[ x=\pm3 \] lie outside the sphere.
Therefore, total number of enclosed charges is \[ 5 \]
Step 3: Calculate the enclosed charge.
Each charge has magnitude \[ q \] Hence, \[ Q_{\text{enclosed}}=5q \]
Step 4: Calculate the electric flux.
Using Gauss's law, \[ \Phi=\frac{Q_{\text{enclosed}}}{\varepsilon_0} \] \[ \Phi=\frac{5q}{\varepsilon_0} \]
Step 5: Final conclusion.
Hence, the electric flux through the spherical surface is \[ \boxed{\frac{5q}{\varepsilon_0}} \]
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