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PHY 102 Test Compilation PHY 102 10 Objective Question
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Question

The mass of a charge-bearing oil droplet in Millikan's apparatus is 10-5 kg. If the droplet has a charge of 7.84 x 10-10 C, determine the electric field strength needed to keep the droplet suspended. (Assume g = 9.8 m/s2)
Options
A 1.25 x 10<sup>4</sup> V/m
B 2 x 10<sup>4</sup> V/m
C 2 x 10<sup>-14</sup> V/m
D 1.25 x 10<sup>5</sup> V/m
Correct Answer
Correct Answer

Option D is the correct answer.

Detailed Explanation

For the droplet to be suspended, the upward electric force (Fe) must balance the downward gravitational force (Fg).Fe = FgqE = mgWhere:q = charge of the droplet = 7.84 x 10-10 CE = electric field strength (what we need to find)m = mass of the droplet = 10-5 kgg = acceleration due to gravity = 9.8 m/s2Rearranging the formula to solve for E:E = mg / qE = (10-5 kg * 9.8 m/s2) / (7.84 x 10-10 C)E = (9.8 x 10-5 N) / (7.84 x 10-10 C)E = 1.25 x 105 V/m

Hint

For a charged particle to be suspended in an electric field, the electric force must exactly counteract the gravitational force. Use the formula F = qE and F = mg.

Question Info

Course: PHY 102
Topic: Test Compilation PHY 102 10
Type: Objective
Difficulty: Not specified
Question 918 of 1819
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