Question

${ \qquad\qquad\huge\underline{{\sf Question}}}$



$\large \texttt{Need help with this one .-.}$
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$\textsf{Two particles have equal masses of 50g each}$$\textsf{and opposite charges}$${\sf + 4.0 +× {10}^{-5} C}$$\textsf{and}$${\sf - 4.0 × {10}^{-5} C.}$

$\textsf{They are released from rest with a separation}$$\textsf{of 1.0 m between them. Find the speeds of }$$\textsf{the particles when the separation is reduced}$$\textsf{to 50 cm}$
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$\textsf{Important note : i already have solved it by }$$\textsf{the method we find force and calculate its}$$\textsf{Acceleration to find the speed, but i didn't}$$\textsf{get the answer from energy conservation }$$\textsf{method somehow, so please use}$

$\textbf{❝ Energy conservation method ❞}$

$\large \textsf{Thanks ! }$​

Answer 1

Hello, law of conservation of energy says sum of kinetic & potential energies at initial point equals to sum of kinetic & potential energies at final point.

please refer the attachment for complete solution!

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Answer 2

54m/s

Solution Given:

Magnitude of charges, q = 4.0 x 10-5 C

Initial separation between charges, r = 1 m

Initial speed = 0; so, initial K.E. = 0

Mass of the particles, m = 50 g =0.05 kg

ReLet 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.

$\frac{1}{4\pi E_o}*\frac{q_1*q_2}{r}=2*\frac{mv^2}{2}+\frac{1}{4\pi E_o}*\frac{q_1*q_2}{r/2}$

$\frac{-1}{4\pi E_o}*\frac{q^2}{r}=mv^2-\frac{2}{4\pi E_o}*\frac{q^2}{r}$

$mv^2=\frac{1}{4\pi E_o}*\frac{q^2}{r}$

$v=\sqrt{\frac{-1}{4\pi E_o}*\frac{q^2}{r}*\frac{1}{m}}$

$v=\sqrt{\frac{9*10^{-9}*(4*10^{-5})^2}{1*0.05}}$

v=16.96m/s

Required velocity of each particle is 17 m/s.