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Consider a positive charge of charge q placed in between plates which are charge with charges opposite in sign and equal in magnitude.A uniform magnetic field is present downwards in the right half of the midway between charged plates then charge slowly accelerates and after reaching the right half,the charge deflects and gets out of the electric field produced due to charged plates but magnetic field is present even outside the plates then charge further deflects and comes in between the plates and it starts accelerating again.this process repeats and the velocity keeps on increasing. Here work done by magnetic field is zero.Charge in the plates remain constant ensuring constant electric field between plates.The ball is gaining kinetic energy continuously.Where does this energy come from?The charged plates are kept between rigid walls to stop movement of plates due to force.The image of this situation is given below.

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The magnetic field may do no work on the test charge, but the electric field does.

I assume you understand what's happening if you only look in between the plates (because there the electric field is constant). If the charge gets closer to the negative plate, it gains (kinetic) energy. If the charge moves further from the negative plate, it loses (kinetic) energy.

Now, when you leave the area between the plates, the specifics of the field are messier, but it's not sharp boundary between a full-strength field inside and zero field outside. The field becomes weaker, but is still present. If the particle can gain $E$ energy by moving between two points between the plates, it will require $E$ energy to move backward even if you pick a path that is outside the plates. The field there might be weaker, but your path will be longer. Together, it will sum to the same value since the electric field is conservative.

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  • $\begingroup$ If the plates are large electric field is very less out side as compared to inside so decrease in velocity outside the plates is less than increase in velocity inside the result is gaining of kinetic energy of ball $\endgroup$ – Sai Charan Reddy May 23 '18 at 17:36
  • $\begingroup$ That's speculation on your part. While the strength does decrease, the direction changes as well. The particle is losing energy over the entire path, not just the perpendicular component. The energy loss is identical over any path you pick. $\endgroup$ – BowlOfRed May 23 '18 at 17:42
  • $\begingroup$ If the strength of electric field is less outside the plates it ensures less energy decrease of charged body outside the plates, how can you say that energy gained by the charged body inside the plates is equal to energy lost by charged body outside the plates. $\endgroup$ – Sai Charan Reddy May 24 '18 at 17:57
  • $\begingroup$ Because that's the way electrostatic fields work. You can prove that any field made from charge distribution (which is how yours is made) is conservative and all paths from A to B involve the same gain/loss of electrical potential energy. You can declare that the field outside the plates is much weaker, but it will really be strong enough that the energy you lose is the same. $\endgroup$ – BowlOfRed May 24 '18 at 18:00
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You're not correct saying that the charge is constantly gaining kinetic energy. Any motion towards the left reduces its kinetic energy because it gains potential energy. Further, if the charge doesn't come in exactly perpendicular to the left plate, then it will be deflected to the right plate at an angle this time, and eventually the charge will collide with the right plate.

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  • $\begingroup$ Loss in potential energy when it travels inside the plates is greater than gain in potential energy as electric field strength is less outside the plates compared to electric field inside the plates. $\endgroup$ – Sai Charan Reddy May 23 '18 at 17:37
  • $\begingroup$ @SaiCharanReddy, all electrostatic fields are conservative. You cannot have different change in potential energy by taking a different path. $\endgroup$ – BowlOfRed May 23 '18 at 17:54
  • $\begingroup$ Yes, but the path the charge takes outside the plates is longer. It works out that the gain in potential energy is the same for any path you pick, as @BowlOfRed explained. $\endgroup$ – HiddenBabel May 23 '18 at 18:03
  • $\begingroup$ Is the field not conservative?How can you justify that the charge takes a longer path outside? $\endgroup$ – Sai Charan Reddy May 24 '18 at 19:12

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