The electric field intensity just sufficient to balance the earth's gravitational attraction on an electron…
- $-5.6 \times 10^{-11} \mathrm{~N} / \mathrm{C}$
- $-4.8 \times 10^{-15} \mathrm{~N} / \mathrm{C}$
- $-1.6 \times 10^{-19} \mathrm{~N} / \mathrm{C}$
- $-3.2 \times 10^{-19} \mathrm{~N} / \mathrm{C}$
Solution
$\overrightarrow{\mathrm{E}}=-\frac{9.1 \times 10^{-31} \times 10}{1.6 \times 10^{-19}}=-5.6 \times 10^{-11} \mathrm{~N} / \mathrm{C}$ ,
Asked in: JEE Mains - Electrostatics - Test 2