Question

q. Two point charges of +16 micro coulomb and -9 micro coulomb are placed 8 cm apart in air .Determine the position of the point at which the resultant field is zero .

Answer 1

Given :

The The magnitude of two charges are = $q_1$  = + 16 $\mu$

                                                                     $q_2$  = - 9 $\mu$c  

The distance between two charges  = 8 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 = ( 8 - 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 $\dfrac{q_2}{(8-x)^{2} }$

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

i.e      k  $\dfrac{q_1 }{x^{2} }$   =   k $\dfrac{q_2}{(8-x)^{2} }$

Or,   $\dfrac{16 }{x^{2} }$    =  $\dfrac{9}{(8-x)^{2} }$

Or,    16 ( 8 - x)² =  9 × x²

O,     16 [ 64 + x² - 16 x ] =  9 × x²

Or,   1024 + 16 × x² - 256 x =  9 × x²

Or,   - 9 × x² + 16 × x² - 256 x + 1024 = 0

Or,    7 × x²  - 256 x + 1024 = 0

now, Solving this quadratic equation

 x = $\dfrac{-(-256)\pm \sqrt{(-256)^{2}-4\times 7\times 1024}}{2\times 7}$

∴       x = 32 cm  , 4.5 cm

So, The distance = x = 4.5 cm

Hence, The force on the charge q will be zero when it is placed at a distance of 4.5 cm   Answer