Question

An infinitely long straight wire with uniform line charge density $\lambda$ lies at a perpendicular distance $d$ from a point O. The total electric flux through the surface of a sphere of radius $R \gt d$ centred at O, is

• A. $\frac{2\lambda}{\epsilon_0}\sqrt{R^2 - d^2}$
• B. $\frac{2\lambda}{\epsilon_0}\sqrt{Rd}$
• C. $\frac{2\lambda}{\epsilon_0}\sqrt{R^2 + d^2}$
• D. $0$

Hint:

For any Gauss's Law problem, focus solely on calculating the enclosed charge.
Geometry of the intersection of a line with a sphere simplifies to a simple 2D circle chord problem.

Answer 1

The Correct Option is A

Step 1: Recall the heart of Gauss's law.
The total electric flux out of any closed surface depends only on the charge sitting inside it, nothing else. The shape of the surface does not matter.

Step 2: Write the law cleanly.
\[ \Phi = \frac{q_{\text{enclosed}}}{\epsilon_0}. \] So we just need the amount of wire charge trapped inside the sphere.

Step 3: Find how much wire is inside.
The wire passes a perpendicular distance $d$ from the centre O. The sphere has radius $R$. The chord of the wire that lies inside the sphere has half-length $\sqrt{R^2 - d^2}$ on each side of the closest point.

Step 4: Get the trapped length.
So the full length of wire inside the sphere is \[ \ell = 2\sqrt{R^2 - d^2}. \]

Step 5: Convert length to charge.
With line charge density $\lambda$, the enclosed charge is \[ q = \lambda \ell = 2\lambda \sqrt{R^2 - d^2}. \]

Step 6: Drop it into Gauss's law.
\[ \Phi = \frac{q}{\epsilon_0} = \frac{2\lambda}{\epsilon_0}\sqrt{R^2 - d^2}. \] \[ \boxed{\Phi = \dfrac{2\lambda}{\epsilon_0}\sqrt{R^2 - d^2}} \]