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two charged particles if mass 4m and m having charges +q and +3q are placed in uniform electric field .they are aloowed to move for two seconds. find the ratio of their kinetic energies . assuming their own electric field intraction to each other is zero.
Since the question is ambiguous, I have assumed that the particles have been placed in the field at rest. We know the Coulombic force on each particle is the product of the charge on the particle and the electric field intensity, F=E*q.
Further, we can assume that the Electric field is due to a constant charge of magnitude Q.
Then, E=kQ/r2, where k is the Coulomb constant 9*109 N m2C-2.
Thus, by calculating the Coulombic force for each particle, we can equate this to force acting on the particle according to Newtons Second Law.
F=ma, F=kqQ/r2 => ma=kqQ/r2. Subsequently we get an expression for acceleration of the body. Now, by assuming that it was initially at rest, we can use equations of 1 dimensional motion to get the expression for velocity after 2 seconds. Using this velocity in the expression for Kinetic Energy, K=mv2/2, we get expressions for Kinetic Energy of both particles. Dividing them and cancelling out the common terms, we get the ratio 1:36. Please refer to the image for detailed calculations.
If my logic is incorrect or there are any errors in calculation or otherwise, please let me know.
In the above reply, please right click the image and select "Open image in new tab" to view the full calculation and the answer.
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