# An infinite number of charges each equal to q are placed along the x-axis at x = 1, x = 2,x = 4,x...

## Question:

An infinite number of charges each equal to q are placed along the x-axis at x = 1, x = 2,x = 4,x = 8 and so on. Find the potential and electric field at the point x = 0 due to this set of charges.

## Electric Field :

The electric force per unit charge or the electric potential per unit length is called the electric field. The SI unit of the electric field is {eq}\text{N/C} {/eq}. The direction of the electric field is from the positive charge to the negative charge.

Electric field due to a charge at any point is formulated as:

$$\color{red}{E=\frac{F}{q}=\frac{Q}{4\pi\varepsilon_0 R}} $$

Here:

- {eq}\varepsilon_0 {/eq} is the permittivity of the vaccum.

- {eq}Q {/eq} is the charge.

- {eq}R {/eq} is the distance of the point from the charge.

## Answer and Explanation:

**Given:**

- The charges {eq}q {/eq} is placed on {eq}x=1,2,4,8 {/eq}.

Now recall for the electric potential:

{eq}\displaystyle{V=\frac{Kq}{R}} {/eq}

So the electric potential at the origin is:

{eq}\begin{align} V&=\frac{Kq}{R_1}+\frac{Kq}{R_2}+\frac{Kq}{R_3}+\frac{Kq}{R_4}\\ &=Kq\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)\\ &=Kq\left(\frac{1}{1-\frac{1}{2}}\right) \ \left(\text{by solving Geometric series} \ s=\frac{a}{1-r}\right)\\ V&=\color{blue}{2Kq} \end{align} {/eq}

Now recall for the electric field:

{eq}\displaystyle{E=\frac{Kq}{R^2}} {/eq}

So the electric field at the origin is is:

{eq}\begin{align} E&=\frac{Kq}{R_1^2}+\frac{Kq}{R_2^2}+\frac{Kq}{R_3^2}+\frac{Kq}{R_4^2}\\ &=Kq\left(1+\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{8^2}\right)\\ &=Kq\left(1+\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}\right)\\ &=Kq\left(\frac{1}{1-\frac{1}{2^2}}\right) \ \left(\text{by solving Geometric series} \ s=\frac{a}{1-r}\right)\\ E&=\color{blue}{\frac{4Kq}{3}} \end{align} {/eq}

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