Question

Two particles of masses 5.0 g each and opposite charges of +4.0 × 10⁻⁵ C and −4.0 × 10⁻⁵ C are released from rest with a separation of 1.0 m between them. Find the speeds of the particles when the separation is reduced to 50 cm.  

Answer 1

Given:
Magnitude of charges, q =  4.0 × 10⁻⁵ C
Initial separation between charges, r = 1 m
Initial speed = 0; so, initial K.E. = 0 
Mass of the particles, m = 5.0 g =0.005 kg
Let the required velocity of each particle be v.
By the law of conservation of energy,
Initial P.E. + Initial K.E. = Final P.E. + Final K.E. 

\[\frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r} = 2 \times \frac{1}{2}m v^2 + \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r/2}\]
\[ \Rightarrow \frac{- 1}{4\pi \epsilon_0}\frac{q^2}{r} = m v^2 - \frac{2}{4\pi \epsilon_0}\frac{q^2}{r}\]
\[ \Rightarrow m v^2 = \frac{1}{4\pi \epsilon_0}\frac{q^2}{r}\]
\[ \Rightarrow v = \sqrt{\frac{1}{4\pi \epsilon_0 m}\frac{q^2}{r}}\]
\[ \Rightarrow v = \sqrt{\frac{9 \times {10}^{- 9} \times \left( 4 \times {10}^{- 5} \right)^2}{0 . 005 \times 1}}\]
\[ \Rightarrow v = 53 . 66\] m/s