Question

Two point charges 4micro C and 1 micro C are separated by a distance of 2 m in air. Find the point on the line joining charges at which the net electric field of the system is zero.​

Answer 1

Answer:

QUESTION

Two point charges 4micro C and 1 micro C are separated by a distance of 2 m in air. Find the point on the line joining charges at which the net electric field of the system is zero.

GIVEN

Q1=4×10^-6C

Q2=1×10^-6C

distance, r=2m

TO FIND

point on the line joining charges at which the net electric field of the system is zero.

• let that point be present at distance x from charge Q1. ......r1
• let that point be present at distance r-x from charge Q2.....r2

FORMULA

electric field,E=KQ/r

K=9×10^9

SOLUTION

• let E1 be the electric field at x due to charge Q1
• let E2 be the electric field at x due to charge Q2

NET ELECTRIC FIELD AT X SHOULD BE ZERO

E1+E2=0

E1=-E2

$\frac{kq1}{ {r1}^{2} } = - \frac{kq2}{ {r2}^{2} } \\ \frac{q1}{ {x}^{2} } = - \frac{q2}{ { ({r - x})^{} }^{2} } \\ \frac{4 \times {10}^{ - 6} }{ {x}^{2} } = - \frac{1 \times {10}^{ - 6} }{ {(r - x)}^{2} } \\ \frac{4}{ {x}^{2} } = - \frac{1}{ {(r - x)}^{2} } \\ \frac{ {(r - x)}^{2} }{ {x}^{2} } = - \frac{1}{4} \\ taking \: square \: root \: on \: both \: sides \: we \: get \\ \frac{r - x}{x} = - \frac{1}{2 } \\ \frac{r}{x} - \frac{x}{x} = - \frac{1}{2} \\ \frac{r}{x} - 1 = - \frac{1}{2} \\ \frac{2}{x} = 1 - \frac{1}{2} \\ \frac{2}{x} = \frac{1}{2} \\ x = 2 \times 2 \\ x = 4m$

ANSWER

the point lies at 4 m from Q1 .

R2=r-x=2-4=|-2|=2 m

the point lies at 2m from Q2

REASON

since both the charges are positive, neutral point lies outside them