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
A
2.11 x 10\(^{14}\)ms\(^{-2}\)
B
2.11 x 10\(^{13}\)ms\(^{-2}\)
C
1.20 x 10\(^{14}\)ms\(^{-2}\)
D
1.20 x 10\(^{13}\)ms\(^{-2}\)
Verified Correct Answer: A.
Physical Principles & Numerical Solution
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\(^{-2}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #70510)