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PHY 102 Test Compilation PHY 102 21 Objective Question
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Question

An electron is positioned in an electric field. The force on the electron due to the electric field is equal to the force of gravity on the electron. What is the magnitude of this electric field?
Options
A 8.93 x 10-30 N/C
B 5.69 x 10-11 N/C
Correct Answer
C 5.58 x 10-9 N/C
D 1.44 x 10-9 N/C
Correct Answer

Option B is the correct answer.

Detailed Explanation

For equilibrium: qE = mg => E = mg/q. E = (9.11x10-31 kg)(9.8 m/s2)/(1.6x10-19 C) = (8.93x10-30)/(1.6x10-19) = 5.58x10-11 N/C ~ 5.69x10-11 N/C.

Hint

For equilibrium: qE = mg => E = mg/q.

Question Info

Course: PHY 102
Topic: Test Compilation PHY 102 21
Type: Objective
Difficulty: Not specified
Question 1293 of 1819
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