Under the action of a given coulombic force the acceleration of an electron is $2.5 \times {10^{22}}{\rm{m}}/{{\rm{s}}^2}.$ Then the magnitude of acceleration of a proton under the action of same force is nearly

$1.5 \times {10^{19}}{\rm{m}}/{{\rm{s}}^2}$

$1.6 \times {10^{ - 19}}{\rm{m}}/{{\rm{s}}^2}$

$9.1 \times {10^{31}}{\rm{m}}/{{\rm{s}}^2}$

$1.6 \times {10^{27}}{\rm{m}}/{{\rm{s}}^2}$

JEE PHYSICS MCQ: ELECTRIC FIELDS AND CHARGES

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Application +4 | -1

Under the action of a given coulombic force the acceleration of an electron is

2.5 \times {10^{22}}\mathrm{m}/{\mathrm{s}^2}.

Then the magnitude of acceleration of a proton under the action of same force is nearly

View answer and explanation

Correct option: C

1.5 \times {10^{19}}\mathrm{m}/{\mathrm{s}^2}

Option C (

1.5 \times {10^{19}}\mathrm{m}/{\mathrm{s}^2}

) is correct because it correctly applies Newton's second law and the relationship between force, mass, and acceleration.

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r.

4N

\left( {{\varepsilon _r} = 8} \right)

N

5 \times {10^{ - 2}}