Question

Two charges of 25uc and 36uc are placed 10cm apart from each other then the force on the charge q will be zero when it is placed at a distance of..?

Answer 1

Answer:

Two charges of 25uc and 36uc are placed 10cm apart from each other then the force on the charge q will be zero when it is placed at a distance of..?

Answer : the force on the charge q will be zero when it is placed at a distance 4.5cm from charge 25μC.

Explanation:

Two charges of 25μC  and 36μC are placed 10cm apart from each other. Suppose a point P, on line the two charges, distance from one charge  is x cm, where electric field intensity becomes zero.

as we know

electric field intensity due to  at Point charge =  $k\frac{q}{r^{2} }$

k(25μC) / x² = k(36μC) / (10 - x)²

⇒ 25 / x² = 36 / (10 - x)²

⇒ 5² / x² = 6² / (10 - x)²

⇒ 5 / x = 6/(10 - x)

⇒ 5(10 - x) = 6x

⇒ 50 = 11x => x = 55 / 11 = 4.5

Therefore the force on the charge q will be zero when it is placed at a distance 4.5cm from charge 25μC.

Answer 2

Given :

The magnitude of two charges are = $q_1$  = 25 $\mu$ c

                                                            $q_2$  = 36 $\mu$ c  

The distance between two charges  = 10 cm

To Find :

The distance at which charge is placed when the force on the charge q will be zero

Solution ;

Let The distance of charge $q_1$ from point p = x cm

So,  The distance of charge $q_2$ from point p = ( 10 - x ) cm

Now,

Electric force at any point at d distance is given as , F = k $\dfrac{q_1 q_2}{d^{2} }$

     where k = 9 × $10^{9}$

So, Force due to charge $q_1$ = k $\dfrac{q_1 }{x^{2} }$

And  Force due to charge $q_2$ = k $\dfrac{q_2 }{(10-x)^{2} }$

And ,    Force due to charge $q_1$  =  Force due to charge $q_2$

i.e      k $\dfrac{q_1 }{x^{2} }$    =   k $\dfrac{q_2 }{(10-x)^{2} }$

Or,     $\dfrac{25}{x^{2} }$  =  $\dfrac{36 }{(10-x)^{2} }$

Or,    25 ( 10 - x)² = 36 × x²

O,     25 [ 100 + x² - 20 x ] = 36 × x²

Or,   2500 + 25 × x² - 200 x = 36 × x²

Or,   36 × x² - 25 × x² + 200 x - 2500 = 0

Or,    9 × x²  + 200 x - 2500 = 0

now, Solving this quadratic equation

        x = $\dfrac{-200\pm \sqrt{{(-200)^{2}}- 4\times 9\times (-2500))} }{2\times 9}$

∴       x = 8.9 cm  , - 31.14 cm

So, The distance = x = 8.9 cm

Hence, The force on the charge q will be zero when it is placed at a distance of 8.9  cm   Answer