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Question: An alpha particle kept in an electric field of strength \(100N{{C}^{-1}}\)will experience a force of...

An alpha particle kept in an electric field of strength 100NC1100N{{C}^{-1}}will experience a force of:
A: 3.2×1017N3.2\times {{10}^{-17}}N
B: 1.6×106N1.6\times {{10}^{-6}}N
C: 3.2×1015N3.2\times {{10}^{15}}N
D: 1.6×1015N1.6\times {{10}^{15}}N

Explanation

Solution

We’re given an alpha particle whose charge is a known value to us. We’re also given the electric field. We can easily find the force experienced by direct substitution into the formula.

Complete Step by step answer:
As we know, F=q×EF=q\times E, where F is the force experienced by the particle, E is the electric field and q is the charge possessed by the particle in the given field.
An electric field is capable of exerting a force on particles inside an electric field. The same instance is given here. We’re given an alpha particle whose charge can be given by
q=+2e=2×1.6×1019=3.2×1019Cq=+2e=2\times 1.6\times {{10}^{-19}}=3.2\times {{10}^{-19}}C
We’re also given the electric field in which the alpha particle resides, that is
E=100NC1E=100N{{C}^{-1}}
We know that the alpha particle in this field will experience a force F=q×EF=q\times E
Hence, on substituting the values. We obtain
 F=3.2×1019×100 F=3.2×1017N \ F=3.2\times {{10}^{-19}}\times 100 \\\ \Rightarrow F=3.2\times {{10}^{-17}}N \\\
Therefore we can conclude that the particle experiences a force of 3.2×1017N3.2\times {{10}^{-17}}N

Hence, option A is the correct answer among the given options.

Note: It should be noted that the direction of the electric field is the direction of that force on a positive charge. The actual force on a particle with charge q is given by F = qE. It points in the opposite direction of the electric field E for a negative charge.