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: Understanding the Question:
The problem asks for the total electric flux passing through a sphere of radius $R$ enclosing a portion of an infinitely long straight charged wire.

Step 2: Key Formula or Approach:
According to Gauss's Law, the total electric flux $\Phi$ through any closed surface is equal to the net charge enclosed by the surface divided by the permittivity of free space:
\[ \Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]
Since the wire has a uniform linear charge density $\lambda$, the enclosed charge is:
\[ Q_{\text{enclosed}} = \lambda L \]
where $L$ is the length of the wire segment that lies inside the sphere.

Step 3: Detailed Explanation:

• The wire is straight and is located at a perpendicular distance $d$ from the center O of the sphere.

• Since $R \gt d$, the wire cuts through the sphere. The intersection of the wire with the sphere forms a chord.

• Let us find the length $L$ of this chord.
- Draw a perpendicular from the center O to the wire. The length of this perpendicular is $d$.
- The distance from the center O to the points where the wire intersects the sphere is equal to the radius $R$.
- This forms a right-angled triangle where the hypotenuse is $R$, one leg is $d$, and the other leg is half the chord length ($L/2$).

• Using the Pythagorean theorem:
\[ \left(\frac{L}{2}\right)^2 + d^2 = R^2 \]
\[ \frac{L}{2} = \sqrt{R^2 - d^2} \implies L = 2\sqrt{R^2 - d^2} \]

• The total charge enclosed by the sphere is:
\[ Q_{\text{enclosed}} = \lambda L = 2\lambda\sqrt{R^2 - d^2} \]

• Using Gauss's Law, the total electric flux through the sphere is:
\[ \Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0} = \frac{2\lambda}{\epsilon_0}\sqrt{R^2 - d^2} \]



Step 4: Final Answer:
The total electric flux is $\frac{2\lambda}{\epsilon_0}\sqrt{R^2 - d^2}$.