Question

Two point charges $8qand-2q$ are located at$x=0andx=l$respectively. The point on the x-axis at which the net electric field is zero due to these charges is

• A. 2L
• B. $\frac{L}{4}$
• C. 4L
• D. 8L

Answer 1

The correct option is A

2L

Step 1: Given Data

Two-point charges $8qand-2q$ are located at$x=0andx=l$respectively.

Step 2: Formula used

Electric Field

When the charge is present in any form, a point in space has an electric field that is connected to it.

The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field.

Consider the region between $x=-\infty andx=0.$ The electric field due to +8q is towards the left and that due to $-2q$towards the right.

A generalized electric field $x=d(<0)isE=\frac{1}{4\pi {\epsilon }_0}\left(-\frac{8q}{d^2}+\frac{2q}{{(L-d)}^2}\right)$

Here, d is a charge and L is a location. q is the quantum.

Step 3: Calculating the charge

If $E=0$ then $-\frac{8q}{d^2}=\frac{2q}{{(L-d)}^2}\Rightarrow \frac{d^2}{{(L-d)}^2}=4$

However, if$d=0$, then$\left|d\right|=|L-d|$. Therefore, there is no answer.

Consider the area between$x=0andx=L$.

Both the electric field caused$+8qandby2q$ is directed rightward.

Therefore, no point in $(0,L)$having an electric field of zero.

Consider the area between$x=Landx=+\infty$.

Due to$+8qand2q$, respectively, the electric field is to the right and to the left.

a generalized electric field $x=d(>L)isE=\frac{1}{4\pi {\epsilon }_0}\left(+\frac{8q}{d^2}+\frac{2q}{{(d-L)}^2}\right)$

If $E=0$ then $-\frac{8q}{d^2}=\frac{2q}{{(d-L)}^2}\Rightarrow \frac{d}{(d-L)}=\pm 2\Rightarrow d=2Lord=\frac{2L}{3}$

But, $d>L$. Hence $d=2L$

Therefore the net electric field is zero for $x=2L$.

Therefore the correct option is A and incorrect options are (B), (C), and (D).