A particle A has charge +q and particle B has charge +4 q with each of them having same mass m . Whe

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2. Electric Potential and Capacitance

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A particle $A$ has charge $+q$ and particle $B$ has charge $+4 q$ with each of them having the same mass $m$. When allowed to fall from rest through the same electric potential difference, the ratio of their speeds $\frac{V_A}{V_B}$ will become

A

$1: 2$

B

$2: 1$

C

$1: 4$

D

$4: 1$

Solution

(a)

$q V=\frac{1}{2} m V_A^2 \quad V_B=\sqrt{\frac{8 q V}{m}}$

$V_A=\sqrt{\frac{2 q V}{m}}$

$\frac{V_A}{V_B}=\frac{1}{2}$

Standard 12

Physics

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