For a point charge \(Q\) in vacuum, a graph is drawn of the magnitude of the electric field \(E\) against \(1/r^{2}\), where \(r\) is the distance from the charge. The graph is a straight line through the origin. The slope of this line is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\frac{Q}{4\pi\varepsilon_0}\)
The field of a point charge decreases as \( 1/r^{2} \). Take \( x=1/r^{2} \) as the quantity on the horizontal axis.
\( \begin{aligned} E &= \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^{2}} \\ &= \left(\frac{Q}{4\pi\varepsilon_0}\right)x \end{aligned} \)
This has the form \( y=mx \), a straight line through the origin. From the above, the slope is \( Q/4\pi\varepsilon_0 \). Hence, a larger charge gives a steeper line.
NCERT (2026-27):
Electric Field – “… The electric field produced by the charge Q at a point r is given as”
\( \displaystyle E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^{2}}\,\hat r \)\( \qquad (1.6) \)