Physics, asked by vikaries1707, 1 year ago

4 equal point charges each of 16 micro coulomb are placed on the 4 corners of the square of side 0.2 m. Calculate the force experienced by any one of the charge

Answers

Answered by abhi178
52

Let ABCD is square of side length 0.2m and four equal point charge each of 16 micro coulomb are placed on the four corner A, B, C and D of the square.

we have to find force experienced by anyone of the charge . we take point A, we have to find force experienced by point A.

to understand see diagram,

force exerted by B on A = Kq²/a²

where q = 16 microcoulomb, a = 0.2m

force exerted by C on A = Kq²/(√2a)² = Kq²/2a²

force exerted by D on A = Kq²/a²

now, resultant of force exerted by B and D on A = √{(Kq²/a²)² + (Kq²/a²)²}= √2Kq²/a² , it is along CA .

now, net force experienced by A = resultant of force exerted by B and D on A + force exerted by C on A

= √2Kq²/a² + Kq²/2a²

= (√2 + 1/2)Kq²/a²

= (2√2 + 1)Kq²/2a²

now, putting k = 9 × 10^9 Nm²/C² , q = 16 × 10^-6 C and a = 0.2 m

= (2√2 + 1) × 9 × 10^9 × (16 × 10^-16)²/2(0.2)²

= (2√2 + 1) × 9 × 256 × 10^-3/(0.08)

= (2√2 + 1) × 9 × 32 × 10^-1

= 110.258 N along CA

Attachments:
Answered by chhayag39
8

Answer:

Explanation:

Let ABCD is square of side length 0.2m and four equal point charge each of 16 micro coulomb are placed on the four corner A, B, C and D of the square.

we have to find force experienced by anyone of the charge . we take point A, we have to find force experienced by point A.

to understand see diagram,

force exerted by B on A = Kq²/a²

where q = 16 microcoulomb, a = 0.2m

force exerted by C on A = Kq²/(√2a)² = Kq²/2a²

force exerted by D on A = Kq²/a²

now, resultant of force exerted by B and D on A = √{(Kq²/a²)² + (Kq²/a²)²}= √2Kq²/a² , it is along CA .

now, net force experienced by A = resultant of force exerted by B and D on A + force exerted by C on A

= √2Kq²/a² + Kq²/2a²

= (√2 + 1/2)Kq²/a²

= (2√2 + 1)Kq²/2a²

now, putting k = 9 × 10^9 Nm²/C² , q = 16 × 10^-6 C and a = 0.2 m

= (2√2 + 1) × 9 × 10^9 × (16 × 10^-16)²/2(0.2)²

= (2√2 + 1) × 9 × 256 × 10^-3/(0.08)

= (2√2 + 1) × 9 × 32 × 10^-1

= 110.258 N along CA

Xd

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