in the given figure calculate the force on a unit positive charge placed at A the length of each side on the square is 4 centimetre
Answers
On the Unit charge kept at the point A,
Net Force will be given by,
Force = Force due to charge at B + Force due to charge at D + Force due to charge at C
Force due to charge at B = kQ₁Q₂/r²
= 9 × 10⁹ × 16 × 10⁻⁹ × 1/(0.04)²
= 9 × 16 × 10000/16
= 90000 N.
Force due to charge at D = 90000 N. [Since, magnitude and distance is same.]
Force due to the charge at C = 9 × 10⁹ × 32 × 10⁻⁹ × 1/(0.04√2)²
= 9 × 16 × 2 × 10000/16 × 1/2
= 90000 × 2/2
= 90000 N.
Now, we have calculated the forces due to all other charges.
Using the Principle of superposition,
Net force on any one charge = Vector sum of all forces on this charge.
Force on A due to B will be in direction of BA. Force on A due to D is in the direction of AD.
Let us first find the net force due to these two.
Fnet = √[(90000)² + (90000)²]
Fnet = 90000√2.
It will bisect BA and AD. Now, Resultant will make an angle of 45 degree with each force.
There is one more force due to C which should also be added vectorially to net force. Force due to C makes an angle 45 with BA.
The total angle between these two forces will be 45 + 45 = 90
Hence, Total force on A = √(Fnet² + Fca²)
= √[(90000√2)² + (90000)²]
= 90000√3 N.
Hope it helps.