Question

Two point charges of +1 µC and +4 µC respectively are placed at 12 cm apart. Find the position of the point

where electric field is zero.

Hey guyz!
pls do help meh out in diz ques.​

Answer 1

HELLO DEAR,

GIVEN:- Two point charges of +1 µC and +4 µC respectively are placed at 12 cm apart.

TO FIND:-  the position of the point where electric field is zero.

SOLUTION:-

Electric field (E) = [1/4π€][q/r²]

Let assume that the electric field of both points charge is zero at point P.

So,

E1 by q1 at point P = E2 by q2 at point P

   

    [1/4π€][q1/x²]  =   [1/4π€][q2/(12-x)²]

${\bold{\frac{9*10^9 *4*10^{-6}}{x^2} =<em> </em>\frac{9*10^9*1*10^{-6}}{(12-x)^2}}}$

=> 4(144 + x² - 24x) = x²

=> 4x² - x² - 96x + 576 = 0

=> 3x² - 96x + 576 = 0

   

By using Shridhacharya method,

            -(b) ± √[(b)² - 4ac]

X = ————————————

                      2a

         -(-96) ± √[(96)² - 4(3)(576)]

X = ———————————————————

                            2(3)

        + 96 ± √[9216 - 6921]

X = ———————————————————

                            6

         96 ± √[2304]

X = ———————————————————

                            6

         96 ± [ 48 ]

X = ———————————————————

                            6

         96 ± 48

X = ——————————

                    6

x = 24 cm  & 8cm.

[x = 24, neglecting because distance between +4µC and +1µC is 12cm]

taking x = 8cm or 0.08m, the position where electrical field is zero toward +1µC from +4µC is (12-8)cm = 4cm or 0.04m

Therefore, at piont P = 8 cm from q1 is point where electric field is zero.

THANKS.

Answer 2

Given:

Two point charges of +1 µC and +4 µC respectively are placed at 12 cm apart.

To find:

Position where the electric field is zero?

Calculation:

• The point where electric field is zero is known as the NEUTRAL POINT.

• Let neutral point be x distance from +1 µC, and will be (12-x) from +4 µC.

So, at the neutral point, the field intensity from the charges will be equal and opposite to one another:

$\rm \therefore \: E1 = E2$

$\rm \implies \: \dfrac{k(1 \times {10}^{ - 6} )}{ {x}^{2} } = \dfrac{k(4 \times {10}^{ - 6}) }{ {(12 - x)}^{2} }$

$\rm \implies \: \dfrac{1}{ {x}^{2} } = \dfrac{4 }{ {(12 - x)}^{2} }$

Take square-root on both sides:

$\rm \implies \: \dfrac{1}{x } = \dfrac{2 }{ 12 - x }$

$\rm \implies \: 2x = 12 - x$

$\rm \implies \: 3x = 12$

$\rm \implies \: x = 4 \: cm$

So, field intensity is zero at 4 cm from +1 µC charge and (12-4) = 8cm from +4 µC charge.