A proton and an electron are placed in a uniform electric field.

The correct answer is Option (1) → They will experience equal electric forces in opposite directions.

The force on a charge in an electric field is \( \vec{F} = q\vec{E} \). The proton (+e) and the electron (−e) have charges of equal magnitude and opposite sign.

    \( F_p = eE \text{ (along } \vec{E}) \)
    \( F_e = eE \text{ (opposite to } \vec{E}) \)

So the forces are equal and opposite, but the accelerations depend on the masses:

    \( \displaystyle a = \frac{eE}{m}, \qquad \frac{a_e}{a_p} = \frac{m_p}{m_e} \approx 1836 \)

Option (2) They will experience same accelerations in the same direction. — Incorrect. The accelerations are in opposite directions, and their magnitudes also differ.

Option (3) They will experience same accelerations in opposite directions. — Incorrect. The electron’s acceleration is about 1836 times larger, as its mass is much smaller.

Option (4) proton will be accelerated while electron will be decelerated. — Incorrect. Speeding up or slowing down depends on the initial velocity; from rest, both speed up.

Note: NTA dropped this question in the Final Answer Key, probably because none of the options was correct as printed. Option (1) originally read “They will experience equal electric forces in the same direction.” It has been changed to “in opposite directions” so that the question has one clear answer.