First slide

Potential energy

Question

A particle A has charge +q and particle I has charge + 4 q with each of them having the same mass m. When allowed to faII from rest through the same electric potential difference, the ratio of their speeds V_A/V_B will become

Easy

A

2 : 1
B

1 : 2
C

1 : 4
D

4 : 1

Solution

Acceleration = (Force/mass)
$∴ a = \frac{F}{m} = \frac{qE}{m}$
So, $a_{A} = 4 a_{B}$
Now, $v^{2} = u^{2} + 2 ax, where u = 0$ and x is same for both particle
$∴ V_{A}^{2} = 2 a_{A} x and V_{B}^{2} = 2 a_{B} x$
or $\frac{V_{A}^{2}}{V_{B}^{2}} = \frac{a_{A}}{a_{B}} = \frac{1}{4}$
or $\frac{V_{A}}{V_{B}} = \frac{1}{2} or V_{A} : V_{B} = 1 : 2$

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