A charged particle q is shot towards another charged particle Q which is fixed, with a speed v. It approaches Q upto a closest distance r and then returns. If q is shot with speed 2v, the closest distance of approach would be

Question

A charged particle q is shot towards another charged particle Q which is fixed, with a speed v. It approaches Q upto a closest distance r and then returns. If q is shot with speed 2v, the closest distance of approach would be (A)  (r/4)  (B)  (r/2)  (C)  2r  (D)  r

Tardigrade Admin · Accepted Answer

(1/2)mv2=(kQq/r1) ...(i) (1/2)m.4v2=(kQq/r2) .. (ii) From Eqs. (i) and (ii), we can say (r1/r2)=4 r2=(r/4)

Q. A charged particle $q$ is shot towards another charged particle $Q$ which is fixed, with a speed $v$ . It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ is shot with speed $2 v$ , the closest distance of approach would be

$\frac{r}{4}$

$\frac{r}{2}$

$2 r$

$r$

$\frac{3}{2} r$

$\frac{1}{2} m v^{2} = \frac{k Qq}{r _{1}}$ ...(i)
$\frac{1}{2} m .4 v^{2} = \frac{k Qq}{r _{2}}$ .. (ii)
From Eqs. (i) and (ii), we can say
$\frac{r _{1}}{r _{2}} = 4$
$r_{2} = \frac{r}{4}$

Q. A charged particle q is shot towards another charged particle Q which is fixed, with a speed v. It approaches Q upto a closest distance r and then returns. If q is shot with speed 2v, the closest distance of approach would be

4r​

2r​

2r

r

23​r

21​mv2=r1​kQq​ ...(i) 21​m.4v2=r2​kQq​ .. (ii) From Eqs. (i) and (ii), we can say r2​r1​​=4 r2​=4r​