First slide

Potential energy

Question

A particle of mass 2 gm and a charge 1 $μ$ C is held at rest on a frictionless horizontal surface at a distance of 1 metre from a fixed charge of 1 mC. If the particle is released it will be repelled, The speed of the particle when it is at a distance of 10 metres from the fixed charge is

Moderate

A

100 m/s
B

90 m/s
C

60 m/s
D

45 m/s

Solution

The change in potential energy of the charges
$\begin{array}{l} = \frac{q_{1} q_{2}}{4 {πε}_{0}} [\frac{1}{r_{1}} - \frac{1}{r_{2}}] \\ = (9 \times 10^{9}) (1 \times 10^{- 6}) (10^{- 3}) [\frac{1}{1} - \frac{1}{10}] \\ = 9 \times \frac{9}{10} = 8 \cdot 1 J \end{array}$
This is equal to the gain is kinetic energy. Let u be the velocity of the particle. Then
$\frac{1}{2} {mv}^{2} = 8 \cdot 1 or \frac{1}{2} \times (2 \times 10^{- 3}) v^{2} = 8 \cdot 1$
Solving, ne get v = 90 m/s.
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