Question

consider a point charge with q is equal to 1.5 into 10 inverse 8 coulomb.a) what is the radius of an equipotential surface having a potential of 30 volt? take V=0 at Infinity. b)are surfaces whose potentials differ by a constant amount (1.0 volt, say) evenly spaced​

Answer 1

Answer:

bhai sabse pehle me kon aaya tha

Answer 2

Answer:

The radius of an equipotential surface is 4.5m.

Explanation:

charge of the point charge is $q=1.5*10^{-8}C$,

Electric field due to the point charge, $E=\frac{kq}{r^2} \\\\ k=\frac{1}{4\pi\epsilon_0}=9*10^9$

The potential due to point charge is given as $V=-\int\limits^r_{\infty }{E} .\, dr =\int\limits^r_{\infty }{\frac{kq}{r^2} } \, dr=\frac{kq}{r}$

Hence  $30=\frac{9*10^9*2.8*10^{-8}}{r}$⇒$r=\frac{9*10^9*2.8*10^{-8}}{30}=4.5m$

• As we saw $V=\frac{kq}{r}$, ⇒$\frac{dV}{dr}=\frac{-kq}{r^2}$,  from this expression we see that potential is changing as inversely proportional to  $r^2$ and its not constant. Hence surfaces whose potential differ by a constant are not evenly spaced.