Question

A vertical electric field of 2×105 N/C just prevents a drop of water of mass 1.00×10-3 kg from

falling. Find the charge on the droplet.​

Answer 1

SoLuTioN :

→ Given:

✏ A vertical electric field just prevents a drop of water from falling.

✏ Magnitude of electric field = 2 × $10^5$ N/C

✏ Mass of water drop = $10^{-3}$ kg

→ To Find:

✏ Charge on water droplet.

→ Concept:

✏ For equilibrium condition of water droplet....

$\star \: \underline{ \boxed{ \bold{ \sf{ \pink{Weight \: Force = Electrical \: Force}}}}} \: \star$

• Weight force = mg
• Electrical force = qE

→ Calculation:

$\rightarrow \sf \: \blue{mg = qE} \\ \\ \rightarrow \sf \: {10}^{ - 3} \times 10 = q \times 2 \times {10}^{5} \\ \\ \rightarrow \sf \: q = \frac{ {10}^{ - 2} }{2 \times {10}^{5} } \\ \\ \rightarrow \: \boxed{ \red{ \sf{q = 500 \: C}}} \: \star$