Electric force accelerates proton, increasing its speed

Steps:

• Electric field points from +10V plate to 0V plate.
• Proton, being positively charged, experiences force F = qE in the field's direction.
• This force causes acceleration a = F/m toward the 0V plate.
• Acceleration adds to the proton's velocity component, raising kinetic energy and speed.

Why C is correct:

• By conservation of energy, the electric field does positive work on the proton moving toward lower potential, increasing KE = (1/2)mv² and thus speed.

Why the others are wrong:

• A: Direction (downwards) depends on plate orientation, not specified without diagram.
• B: Upwards motion would require attraction to +10V plate, but proton repels from it.
• D: Speed decreases only if moving toward higher potential; here, field accelerates toward lower potential.

Final answer: C