- Home
- Standard 12
- Physics
1. Electric Charges and Fields
medium
A charged water drop whose radius is $0.1\,\mu m$ is in equilibrium in an electric field. If charge on it is equal to charge of an electron, then intensity of electric field will be.......$N/C$ $(g = 10\,m{s^{ - 1}})$
A
$1.61$
B
$26.2$
C
$262$
D
$1610$
Solution
(c) In balance condition $QE = mg = \left( {\frac{4}{3}\pi {r^3}\rho } \right)\,g$
$==>$ $E = \frac{{4 \times (3.14)\,{{(0.1 \times {{10}^{ – 6}})}^3} \times {{10}^3} \times 10}}{{3 \times 1.6 \times {{10}^{ – 19}}}}$$ = 262\,N/C$
Standard 12
Physics
Similar Questions
normal
hard