A particle of mass m=10g and charge q=-6*10^-6 C is released from rest in a regi

Question

A particle of mass m=10g and charge q=-6*10^-6 C is released from rest in a region of non-uniform potential as shown in the diagram below. The particle begins at the point labeled x on the 4 V equipotential. (Ignore effects due to gravity).

What will be the speed of the particle when it crosses the next equipotential line (if the particle would never cross one of the other equipotential lines, indicate that)?

Explanation / Answer

the initial value of potential energy of he particle is P1 = 4*q where q= charge of the particle. now since the charge is -ve the charge moves towards higher potential and thus it would cross the 5V potential. when at the %V potential, the potential energy of the particle is 5*q. now from the principle of conservation of energy the change in potential energy =change in kinetic energy. => 4q-5q = 1/*m*v^2. where v= velocity of the particle when it reaches 5V potential. given q= -6*10^-6C m= 10g. => 6*10^-6 = 1/2*(.01)*v^2. =>v= 3.4*10^-2 m/s => the velocity of the article is v =3.4 *10^-2 m/s.

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