Question

Two point charges $+10\text{nC}$ and $-10\text{nC}$ are separated by a distance of $6\text{cm}$. Find the electric flux hrough a circle of radius $4\text{cm}$ placed with its centre coinciding with the mid - point of line joining the two charges in the perpendicular plane as shown in the figure.

• A. $542.2{\text{N-m}}^2/\text{C}$
• B. $742{\text{N-m}}^2/\text{C}$
• C. $642{\text{N-m}}^2/\text{C}$
• D. $442.4{\text{N-m}}^2/\text{C}$

Answer 1

The correct option is D $442.4{\text{N-m}}^2/\text{C}$

This circle can be assumed as base of a cone having semi - vertex angle $\theta$ as shown in the figure.

From geometry,

$\tan \theta =\frac{4}{3}$

$\Rightarrow \theta ={53}^{\circ }$


Here, flux due to $+10\text{nC}$ and $-10\text{nC}$ will be in the same direction through the given circle.

$\therefore {\varphi }_{net}=2\times \frac{q}{2{ϵ}_0}(1-\cos \theta )=\frac{q}{{ϵ}_0}(1-\cos \theta )$

$=\frac{10\times {10}^{-9}}{9\times {10}^{-12}}(1-\cos {53}^{\circ })$

$=442.4{\text{N-m}}^2/\text{C}$