Question

Two charged particles placed at a separation of 20 cm exert 20 N of Coulomb force on each other. What will be the force if the separation is increased to 25 cm ?

Concept of Physics - 1 , HC VERMA , Chapter "The Force".

Answer 1

Hello Dear.

Given ⇒
Force exerted between the charges (F) = 20 N.
Distance between the Charges (r₁) = 20 cm.
 = 0.2 m

Let the masses of the two charges be m₁ and m₂ respectively.

Now, Using the Formula,

F = (G m₁ m₂) ÷ (r₁)²
(G m₁ m₂) = 20 × (0.2)²
G m₁ m₂ = 0.8  ----eq(i)

Now, If the Separation between the Charges will increased to the 25 cm (or 0.25 m)

Then,
F = (G m₁ m₂) ÷ (0.25)²
 F = 0.8/0.0625    [From eq (i)]
 F = 12.8 N.


Hence, the Force exerted by the charged particles on each other when they are placed at the distance of 25 cm is 12.8 N.


Hope it helps.

Answer 2

HEY!!

_____________________________

✴Two charged particles placed at a separation of 20 cm exert 20 N of Coulomb force on each other.

So, 

⚫F1= 1 / 4π∈0 · q^2 / r2 1

Also, 

⚫F2=14π∈0·q2r22

We have:-

⚫F2 / F1=r2 1/r2 2       

⚫20×20 / 25×25 =16 / 25

⚫F2=16 / 25 ×F1

⚫F2=16 / 25×20 

⚫F2=12.8 N ≈13.0 N

▶▶Therefore, the two charged particles will exert a force of 13.0 N on each other, if the separation is increased to 25 cm.