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NEET UG

Practice

Electrostatics

Electric Charge

Quick practice

Question 1 of 5

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 Å apart is \left(\mathrm{m}_{\mathrm{e}}=9 \times 10^{-31} \mathrm{~kg}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right)

\left(\ { Take }=\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}\right)

A

10^{25} \mathrm{~m} / \mathrm{s}^{2}

B

10^{24} \mathrm{~m} / \mathrm{s}^{2}

C

10^{23} \mathrm{~m} / \mathrm{s}^{2}

D

10^{22} \mathrm{~m} / \mathrm{s}^{2}

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