Question

two point charges Q1 and Q2 exert a force f when kept to certain distance apart.if the charge on each particle is halved and distance between the two particles is doubled then the new force between two particles would be A. f/2B. f/4C. f/8D. f/16I'm getting f/8 but ans is f/16 can anyone tell where I'm going wrong??

Answer 1

f/ 16 is indeed the correct answer dear.



Explanation:


Given:
Charges = q1 and q2.
let the distance be r


Now the f will be
$k \frac{q1q2}{ {r}^{2} }$



The new charges = q1/2 and q2/2
new distance= 2r

Now the f =
$k \frac{ \frac{q1}{2} \times \frac{q2}{2} }{ {(2r)}^{2} } \\ \\ = k \frac{ \frac{q1q2}{4} }{4 {r}^{2} } \\ = k \frac{q1q2}{16 {r}^{2} }$

which is
1/16 times previous force so the new force is 1/16 f.

Answer 2

Answer:The new force between 2 particles would be 1/16 times than the original force.

Explanation:

We know that the force between 2 charged particles is given by,

F = k x  Q₁Q₂/r²

Hence the force between them is directly proportional to product of the charges and inversely proportional to the square of the distance between them,

when the charges becomes half and distance doubled, the new force will be given by,

F' = k x (Q₁/2)(Q₂/2)/(2r)₂

= k x Q₁Q₂/r² x 1/16

=> F' = F/16

Hence the force becomes 1/16 the of the original force.