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PHY 102 Test Compilation PHY 102 19 Objective Question
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Question

When a charge is accelerated through a potential difference of 500V, its kinetic energy increases from 2.0×10⁻⁵J to 6.0×10⁻⁵J. What is the magnitude of the charge?
Options
A 4.0 × 10⁻⁸C
B 8.0 × 10⁻⁸C
Correct Answer
C 1.2 × 10⁻⁷C
D 1.6 × 10⁻⁷C
Correct Answer

Option B is the correct answer.

Detailed Explanation

The work-energy theorem gives q = ΔKE/V = (6.0×10⁻⁵ - 2.0×10⁻⁵)/500 = 4.0×10⁻⁵/500 = 8.0×10⁻⁸C.

Hint

Use q = ΔKE/V, where ΔKE is the change in kinetic energy.

Question Info

Course: PHY 102
Topic: Test Compilation PHY 102 19
Type: Objective
Difficulty: Not specified
Question 1205 of 1819
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