Answer:
Force on proton at point A, \[\mathbf{F=e}{{\mathbf{E}}_{\mathbf{A}}}\mathbf{=1}\mathbf{.6\times 1}{{\mathbf{0}}^{\mathbf{-19}}}\mathbf{\times 40=6}\mathbf{.4\times 1}{{\mathbf{0}}^{\mathbf{-18}}}\mathbf{N}\mathbf{.}\] As the separation between the lines of force at point B is twice of that at point A, so \[{{E}_{B}}=\frac{1}{2}{{E}_{A}}=\frac{1}{2}\times 40=20N{{C}^{-1}}\].
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