Question

q is a positive charge and Q is a conductor. Electric field between charge and conductor is shown in the figure. Field along y-axis is given by $E=E_0(1-k{y}^2)$, where k=0.25. If potential at charge q is taken 100 V, find potential at the point B on conductor as shown in figure. Given $E_0=50N/C$

Answer 1

P.D $\Delta v=-\int E.dy=-\int E_0(1-k{y}^2)dy$ since surface of conductor is equipotential we find potential at A(0,2) instead of B(2,4)

$\Delta V=-E_0{[y-\frac{y^3}{3}k]}_0^2=-50[2-\frac{8}{4\times 3}]$

$\Delta V=-66.67\Rightarrow V_Q-V_p=-66.67$

$V_R-100=-66.67$

$V_R=V_A=V+b=33.33V$