Question

two fixed point charges
4uC and 1uC are separated
from each
by 1m
in air. Find the point on the
line joining charges where resultant electric intensity is zero​

Answer 1

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✤ Required Answer:

Let the point charge on the line joining the two points A(4Q) and B(Q) be P.

• At Point P, the forces are balanced.
• So, electrostatic force exerted by A will be balanced by electrostatic force exerted by B, then only Net force will be 0.
• Also there directions are opposite.

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✤ How to solve?

We know that, the electrostatic force that each of two point charges at a distance r apart exerts on one another is expressed as:

$\large{ \boxed{ \rm{f = k \frac{q1q2}{ {r}^{2} } }}}$

So, by using let's solve the question.....

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✤ Solution:

We know that, the electrostatic forces are balanced. The directions are already equal,so we need to find the magnitudes which are equal.

$\large{ \rm{ \rightarrow \: k \dfrac{4q \times charge \: at \: p}{ {x}^{2} } = k \dfrac{q \times charge \: at \: p}{(1 - x) {}^{2} } }}$

k and Charge at P cancels,

$\large{ \rm{ \rightarrow{ \dfrac{4}{ {x}^{2} } = \dfrac{1}{(1 - x) {}^{2} }}}}$

Cross multiplying,

$\large{ \rm{ \rightarrow{4(1 - x) {}^{2} = {x}^{2} }}}$

$\large{ \rm{ \rightarrow{4(1 + {x}^{2} - 2x) = {x}^{2} }}}$

$\large{ \rm{ \rightarrow{4 + 4 {x}^{2} - 8x = {x}^{2} }}}$

$\large{ \rm{ \rightarrow{4 {x}^{2} - 8x + 4 - {x}^{2} = 0}}}$

$\large{ \rm{ \rightarrow{ 3 {x}^{2} - 8x + 4 = 0}}}$

$\large{ \rm{ \rightarrow{3 {x}^{2} - 6x - 2x + 4 = 0}}}$

$\large{ \rm{ \rightarrow{3x(x - 2) - 2( x - 2) = 0}}}$

$\large{ \rm{ \rightarrow{ (3x - 2)(x - 2) = 0}}}$

$\large{ \rm{ \rightarrow{so \: x = \dfrac{2}{3}m \: or \: 2m}}}$

As the distance between them is 1 m, x can't be equals to 2 m. So x = 2/3 m

☀️ So, Distance between the points:

• At a distance x = 2/3 m from point A.
• At a dis. 1 - x = 1/3 m from point B.

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