Two point charges (+Q) and (-2Q) are placed a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the point where the electric field is zero be at a distance $x$ from the origin on the negative $X$-axis. Let this distance be $d = |x|$.The e

Vedclass · Accepted Answer

$x = -\sqrt{2} a$

Vedclass · Answer

(B) Let the point where the electric field is zero be at a distance $x$ from the origin on the negative $X$-axis. Let this distance be $d = |x|$.The electric field due to charge $+Q$ at distance $a$ from the origin at point $P$ (distance $d$ from origin) is $E_1 = \frac{kQ}{(a+d)^2}$ directed towards the left.The electric field due to charge $-2Q$ at distance $2a$ from the origin at point $P$ is $E_2 = \frac{k(2Q)}{(2a+d)^2}$ directed towards the right.For the net electric field to be zero,$E_1 = E_2$.$\frac{kQ}{(a+d)^2} = \frac{2kQ}{(2a+d)^2}$$\frac{1}{(a+d)^2} = \frac{2}{(2a+d)^2}$Taking the square root on both sides: $\frac{1}{a+d} = \frac{\sqrt{2}}{2a+d}$$2a + d = \sqrt{2}a + \sqrt{2}d$$d(\sqrt{2} - 1) = a(2 - \sqrt{2})$$d = a \frac{\sqrt{2}(\sqrt{2} - 1)}{\sqrt{2} - 1} = \sqrt{2}a$Since the point is on the negative $X$-axis,the coordinate is $x = -\sqrt{2}a$.

Two point charges $(+Q)$ and $(-2Q)$ are placed at distances $a$ and $2a$ respectively from the origin on the $X$-axis. At what distance from the origin is the electric field zero?

Similar Questions

Electric field strength due to a point charge of $5\,\mu C$ at a distance of $80\, cm$ from the charge is:

What is called an electric field?

The unit $Volt/metre$ is the unit of:

Is electric field a scalar or a vector quantity? Why?

Vedclass Test Series

Exam Paper Generator

Online Exam Module