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. \(\dfrac{5q}{\varepsilon_0}\)
• B. \(\dfrac{8q}{\varepsilon_0}\)
• C. \(0\)
• D. \(\infty\)

Hint:

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.

Answer 1

The Correct Option is A

Step 1: State Gauss's Law.
According to Gauss's Law, the total electric flux through any closed surface equals the total charge enclosed divided by the permittivity of free space: \[ \Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \]
Step 2: Identify the positions of all charges.
Charges of magnitude $q$ are placed on the $x$-axis at the origin and at every $1\,\text{cm}$ in both directions. So charges are at $x = 0, \pm 1, \pm 2, \pm 3, \ldots\,\text{cm}$.
Step 3: Determine which charges lie inside the sphere.
The spherical Gaussian surface has radius $2.5\,\text{cm}$ centered at the origin. A charge at position $x$ is inside the sphere if $|x| < 2.5\,\text{cm}$. Checking each position: $x = 0$ (inside), $x = \pm 1\,\text{cm}$ (inside), $x = \pm 2\,\text{cm}$ (inside), $x = \pm 3\,\text{cm}$ (outside, since $3 > 2.5$).
Step 4: Count the enclosed charges.
Charges inside: $x = 0$ (1 charge) $+ x = +1$ (1 charge) $+ x = -1$ (1 charge) $+ x = +2$ (1 charge) $+ x = -2$ (1 charge) $= 5$ charges in total. So $Q_{\text{enclosed}} = 5q$.
Step 5: Calculate the electric flux.
\[ \Phi = \frac{5q}{\varepsilon_0} \]
Step 6: State the conclusion.
Since all charges are positive and identical, the total charge inside the sphere of radius $2.5\,\text{cm}$ is $5q$. The total electric flux through the spherical surface is: \[ \boxed{\Phi = \frac{5q}{\varepsilon_0}} \]