Correct Answer: D) r/4
The closest distance of approach is determined by the principle of conservation of energy. When the charged particle qq is shot towards the fixed charged particle QQ, its initial kinetic energy is converted into electric potential energy at the closest distance of approach rr, where the relative velocity of qq momentarily becomes zero.
1. Initial Energy Conservation (Speed = vv):
• Initial kinetic energy of qq:
KEinitial=12mv2KE_{\text{initial}} = \frac{1}{2}mv^2
• Electric potential energy at the closest distance rr:
PEclosest=kqQrPE_{\text{closest}} = \frac{kqQ}{r}
(Here kk is the Coulomb constant.)
From energy conservation:
12mv2=kqQr(1)\frac{1}{2}mv^2 = \frac{kqQ}{r} \tag{1}
2. Doubling the Speed (v→2vv \to 2v):
If the initial speed of qq is doubled to 2v2v, the initial kinetic energy becomes:
KEinitial (new)=12m(2v)2=2mv2KE_{\text{initial (new)}} = \frac{1}{2}m(2v)^2 = 2mv^2
Let the new closest distance of approach be r′r'. At this point:
2mv2=kqQr′(2)2mv^2 = \frac{kqQ}{r'} \tag{2}
3. Relation Between rr and r′r':
From equations (1) and (2):
kqQr=12mv2andkqQr′=2mv2\frac{kqQ}{r} = \frac{1}{2}mv^2 \quad \text{and} \quad \frac{kqQ}{r'} = 2mv^2
Dividing the two equations:
r′r=kqQ2mv2kqQmv2=14\frac{r'}{r} = \frac{\frac{kqQ}{2mv^2}}{\frac{kqQ}{mv^2}} = \frac{1}{4}
Thus, r′=r4r' = \frac{r}{4}.
The closest distance of approach when the speed is doubled is r/4r/4.
