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

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