Question

Figure shows equipotential surfaces concentric at 'O' the magnitude of electric fields at distance r(in meter) measured from

• A. $\frac{9}{r^2}\left(V{m}^{-1}\right)$
• B. $\frac{6}{r^2}\left(V{m}^{-1}\right)$
• C. $\frac{2}{r^2}\left(V{m}^{-1}\right)$
• D. $\frac{16}{r^2}\left(V{m}^{-1}\right)$

Answer 1

The correct option is B $\frac{6}{r^2}\left(V{m}^{-1}\right)$

We know

$E+\frac{v}{r}$ where $v=\frac{kQ}{r^2}$

from fig

$E_1^1=60/0.1=600v/mr=0.1m$

$E_1=\frac{kQ}{\left(0.01\right)}$ or $kQ=(600\times 0.01)=6$

${11}^{th}{E}_2=\frac{30}{0.2}=150V/mr=0.2m$

$E_2=\frac{kQ}{(0.2{)}^2}kQ=150\times 0.04=6$

$\therefore E_1,E_2,E_3$ can be expressed as

$E_1=\frac{b}{(0.1{)}^2}{E}_2=\frac{b}{(0.2{)}^2}{E}_2=\frac{b}{(0.3{)}^3}-------E_2=\frac{b}{r^2}v/m$