Question

Along the X-axis, three charges$\frac{q}{2},-q$ and $\frac{q}{2}$ are placed at x = 0, x = a and x = 2a respectively. The resultant electric potential at x = a+r (if, a << r) is :

( ${ϵ}_0$ is the permittivity of free space)

• A. $\frac{qa}{4\pi {ϵ}_0{r}^2}$
• B. $\frac{q{a}^2}{4\pi {ϵ}_0{r}^3}$
• C. $\frac{q\left(a^2/4\right)}{4\pi {ϵ}_0{r}^3}$
• D. $\frac{q}{4\pi {ϵ}_{0}r}$

Answer 1

Answer: (B)

$\frac{q{a}^2}{4\pi {ϵ}_0{r}^3}$

Electric potential

$V=\frac{1}{4\pi {ϵ}_o}\frac{q}{r}=\frac{kq}{r}$

potential at (a+r) is the sum of the potentials of all three charges at that point

$V_R=\frac{kq}{2(a+r-0)}-\frac{kq}{a+r-a}+\frac{kq}{2\left(\right(a+r)-2a)}$

$V_R=kq\left(\frac{r(r-a)-2(r-a)(a+r)+r(a+r)}{2(a+r)(r-a)r}\right)=kq\frac{a^2}{r^3-a^{2}r}$

as the assumption is $a<<r$ we can neglect $a$

$V=\frac{kq{a}^2}{r^3}=\frac{1}{4\pi {ϵ}_o}\frac{q{a}^2}{r^3}$