OnlineCBT Logo OnlineCBT

Physics · JEE | Electric Charges and Fields

हिन्दी

What is the acceleration of an electron when subjected to an electric force of magnitude \( 2.3 \times 10^{-8}\,\text{N} \) due to a proton? (Take \( m_e = 9.11 \times 10^{-31}\,\text{kg} \))

  1. A
    \( 1.4 \times 10^{19}\text{ m/s}^2 \)
  2. B
    \( 3.0 \times 10^8\text{ m/s}^2 \)
  3. C
    \( 9.11 \times 10^{15}\text{ m/s}^2 \)
  4. D
    \( 2.5 \times 10^{22}\text{ m/s}^2 \)
View Answer & Solution

Correct answer · Option D

\( 2.5 \times 10^{22}\text{ m/s}^2 \)

Explanation

Using Newton's second law, \( a = \frac{F}{m_e} = \frac{2.3 \times 10^{-8}}{9.11 \times 10^{-31}} = 2.5 \times 10^{22}\,\text{m/s}^2 \).
Address Contact

+91 9694780911

E-mail

onlinecbtportal@gmail.com

Helpline Number

+91 9694780911


Click To Download
Get it on Google Play

ऐप डाउनलोड करने
के लिए Google Play पर
उपलब्ध है

WhatsApp Chat