A vertical electric field of magnitude 4.9 × 105 N/C just prevents a water droplet of a mass of 0.1 g from falling. The value charge on the droplet will be(Given g = 9.8 m/s2)
1.6 × 10–9 C
2.0 × 10–9 C
3.2 × 10–9 C
0.5 × 10–9 C
The Correct Option is B
Solution and Explanation
Since the droplet is at rest
⇒ Net force = 0
⇒ mg = qE
⇒ q=\(\frac{mg}{E}\)=2×10-9 C
The correct option is (B) : 2.0 × 10–9 C
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).