Physics, asked by atul16oct, 11 months ago

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​

Attachments:

Answers

Answered by tiwaavi
17

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.


atul16oct: thanq so much
Similar questions