Question

two point charges +4mc and +1mc are separatedby a distance of 2m in air. find the point on the line joining charges at which the net electric field of the system is zero​

Answer 1

Answer:

Given:

4 mC and 1 mC is separated by a distance of 2 m in air.

To find:

Position where the electric Field intensity will be zero.

Concept:

In the line joining the charges, we will get a point where the field intensity due to both the charges will be equal and opposite. Such that the vectors will cancel each other. This point is called Neutral Point.

Calculation:

∴ E1 = E2

$= > \dfrac{k(4 \times {10}^{ - 3} )}{ {x}^{2} } = \dfrac{k(1 \times {10}^{ - 3} )}{ {(2 - x)}^{2} }$

$= > \dfrac{4}{ {x}^{2} } = \dfrac{1}{ {(2 - x)}^{2} }$

Taking square root on both sides :

$= > \dfrac{2}{ x } = \dfrac{1}{ (2 - x) }$

Transposing to opposite sides :

$= > 4 - 2x = x$

$= > 3x = 4$

$= > x = \dfrac{4}{3} \: m$

So the neutral point is 4/3 m from 1 mC charge and 2/3 m from 4 mC charge

Answer 2

Given

point charges

+4$\mu C$=4×$10^-16$C

+1$\mu C$=1×$10^-16$C

D=2m

To find

Point where net electric field is zero

Solution

let the electric field of the system be S

AS=r

At S

$E_1=E_2$

V$\because$ $\frac{Q_2 \times 1}{4πE_0 r^2}$=$\frac{Q_1 \times 1}{4πE_0 (2-r)^2}$

=>$\frac{4}{r^2}$=$\sf\frac{Q_2 \times 1}{4πE_0 r^2}$

=>$\frac\sqrt{4}}{r^2}}$=$\frac{1}{2-r}$

=>$\frac{2}{r}$=$\frac{1}{2-r}$

=>2(2-r)=r

=>4-2r=r

=>4=3r

$\because$r=$\frac{4}{3}$