The two charges of magnitude 1 μC and 2 μC are separated by 10 cm from each other. At what point on the line joining the two charges is the electric field strength is zero.

Solution:

Given:

Magnitude of charge 1             =       q₁      =     1 μC  = 10^-6 C

Magnitude of charge 2             =       q₂      =     2 μC  = 2 X 10^-6 C

Distance between them           =       r        =      10 cm = 0.1 m

The point P must lie between charges because the force exerted by the charges on a test charge opposes each other.

Let,

E₁ be the electric field at point P by charge q₁.

E₂ be the electric field at point p by charge q₂.

Therefore,

E₁ = E₂

q₁ / ( 4 π ε_o x² ) = q₂ / [ 4 π ε_o ( r - x)² ]

where,

x is a distance of the said point from charge q₁, and (r-x) a distance of the same point from charge q₂

{ q₁ / ( x² ) } = { q₂ / ( r – x )² }

Putting values of q₁ , q₂ and r we get,

{ 10^-6  / ( x² ) } = { ( 2 X 10^-6 ) / ( 0.1 – x )² }

{ 1  / ( x ) } = { (√2  ) / (0.1 - x) }

Taking cross multiplication we get,

( 0.1 – x ) =  ( √2 ) x

0.1   =  ( √2 ) x + x

0.1   =  [ ( √2 ) + 1 ] x

0.1   =  2.4142 x

X= 0.1 / 2.4142

X = 0.04142 m

i.e.

x = 4.142 cm

at 4.142 cm from charge q1 we get a point where is the electric field is ZERO