Answer:
As given, the separation between electric field lines at point A is half of that at point B. Therefore, \[{{E}_{A}}=2{{E}_{B}}\Rightarrow {{E}_{B}}=\frac{{{E}_{A}}}{2}\] \[\therefore \] \[{{E}_{B}}=\frac{60}{2}=30N/C\] We know, force on a charge in a electric field, F = qE. \[\therefore \] \[{{F}_{B}}={{E}_{B}}q=1.6\times {{10}^{-19}}\times 30\] \[=4.8\times {{10}^{-18}}N\]
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