The charge of magnitude 1.6 x 10 \(^{-19}\)C is placed in a uniform electric field of intensity 1200Vm\(^{-1}\). Calculate its acceleration, if the mass of the charge is 9.1 x 10\(^{-31}\)kg
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2.11 x 10\(^{14}\)ms\(^{-2}\)
- 2.11 x 10\(^{13}\)ms\(^{-2}\)
- 1.20 x 10\(^{14}\)ms\(^{-2}\)
- 1.20 x 10\(^{13}\)ms\(^{-2}\)
The correct answer is: A
Explanation
Electric field intensity (E) = \(\frac{\text{force}}{\text{charge}}\) = \(\frac{\text{F}}{\text{q}}\)
Given: E = 1200Vm\(^{-1}\), q = 1.6 x 10 \(^{-19}\)C, m = 9.1 x 10\(^{-31}\)kg
F = E x q
F = 1.6 x 10\(^{-19}\) x 1200 = 1.92 x 10\(^{-16}\)N
F = ma
a = \(\frac{F}{m}\) = \(\frac{1.92\times 10^{-16}}{9.1\times 10^{-31}}\) \(\approxeq\) 2.11 x 10\(^{14}\)ms\(^{-2}\)
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