the two charges of magnitude 1 microcoulomb and 2 microcoulomb 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
Answers
(1/1+roor2)×10 cm from the side of 1 Microcoulamb
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...
Sorry for the calculation part...you can solve it own once you know the concept...hope your answer matches mine....