Question

Two charges 2 microcoulomb and minus 2 microcoulomb are placed at points a and b 5 cm apart

Answer 1

Let q1 = 1 microvoulomb = 1 × 10^-6 C

q2 = 2 microcoulomb = 2 ×10^-6C

seperation between them

r = 10 cm = 10 × 10^-2 m = 10^-1 m

Let A and B be the location where the charges q1 and q2 are placed...and let point P in between the the joining A aand B so that AP = x and BP = r - x ( since AB = r )

Now at Point P , Electric field intensity is zero , this is only if the magnitude of Electric field due to point charges q1 and q2 at P are equal in magnitude but opposite in direction...

We know that , Electric field intesity due to a point charge at location r is E = k Q / r^2 where k = coulomb's constant

then

Electric field due to charge q1 at P

= E1 = k × q1 / ( x ^2)

Electric field due to charge q2 at P

=E2 = k × q2 / ( { r - x } )^2

Clearly E1 = E2

=> k × q1 / ( x ^2) = k × q2 / ( { r - x } )^2

=> q1 / ( x ^2) = q2 / ( { r - x } )^2

=> 1 × 10 ^-6 / ( x ^2) =2 × 10^-6 / ( { r - x } )^2

=> 1 / ( x ^2) =2 / ( { r - x } )^2

On taking reciprocal on both sides , we get

x ^2 = ( r - x ) ^2 / 2

Taking square root gives

x = r - x / root ( 2 )

=> root ( 2 ) x = r - x

=> root ( 2 ) x + x = r

=> x [ root ( 2 ) + 1 ] = r

=> x = r / root (2 ) + 1

=> x = r / (root (2 ) + 1 ) ×{ root (2 ) - 1} /{root (2 ) - 1 } [ Rationalisation ]

=> x = r ( root (2 ) - 1 ) / ( 2 - 1)

=> x = ( 10 ^-1 ) ( 1.41 - 1 ) =( 1 / 10 ) × 0.41

x = 0.041m = 0.041 m × 100 cm / 1m = 4.1 cm

=> x = 4.1 cm...

Thus E will be zero at 4.1cm from the q1 at a point P...