Question

The potential and electric field in the above set up if the consecutive charges have opposite sign is :

• A. $\frac{q}{6\pi {ϵ}_0}$ and $\frac{q}{5\pi {ϵ}_0}$
• B. $\frac{q}{6\pi {ϵ}_0}$ and $\frac{q}{6\pi {ϵ}_0}$
• C. $\frac{q}{5\pi {ϵ}_0}$ and $\frac{q}{6\pi {ϵ}_0}$
• D. $\frac{q}{4\pi {ϵ}_0}$ and $\frac{q}{6\pi {ϵ}_0}$

Answer 1

The correct option is A $\frac{q}{6\pi {ϵ}_0}$ and $\frac{q}{5\pi {ϵ}_0}$

$\Rightarrow V=\frac{kq}{1}-\frac{kq}{2}+\frac{kq}{2^2}-\frac{kq}{2^3}+----$

$V_f=kq\left(\right(1+\frac{1}{2^2}+\frac{1}{2^4}----)-(\frac{1}{2}+\frac{1}{2^3}+---\left)\right)$

$=\frac{q}{4\pi {ϵ}_0}\left(\right(\frac{1}{1-\frac{1}{2^2}})-(\frac{{}^1{/}_2}{1-\frac{1}{2^2}}\left)\right)$

$V_f=\frac{q}{4\pi {ϵ}_0}(\frac{4}{3}-\frac{2}{3})=\frac{q}{6\pi {ϵ}_0}$

$E=\frac{ka}{1^2}-\frac{kq}{(2{)}^2}+\frac{kq}{(2^2{)}^2}-\frac{kq}{(2^3{)}^2}$

$E=kq(\frac{1}{1}+\frac{1}{2^4}+\frac{1}{2^8}+--)-(\frac{1}{2^2}+\frac{1}{2^6}+---)$

$E=\frac{q}{4\pi {ϵ}_0}\left(\right(\frac{1}{1-\frac{1}{16}})-\frac{\frac{1}{4}}{1-\frac{1}{16}})$
$E=\frac{q}{5\pi {ϵ}_0}$