Explanation
The dimension of the magnetic field intensity B is
The electrostatic force is given as,
$$F = \dfrac{{{Q_1}{Q_2}}}{{4\pi {\varepsilon _0}{R^2}}}$$
$${\varepsilon _0} = \dfrac{{{Q_1}{Q_2}}}{{4\pi F{R^2}}}$$
Dimensions of $${\varepsilon _0}$$ are,
$$ = \dfrac{{{A^2}{T^2}}}{{ML{T^{ - 2}} \times {L^2}}}$$
$$ = \dfrac{{{A^2}{T^4}}}{{M{L^3}}}$$
$$ = {M^{ - 1}}{L^{ - 3}}{T^4}{A^2}$$
Dimension, $$\dfrac{a}{{{V}^{2}}}$$ is the same as pressure dimension is $$N{{m}^{-2}}$$
$$ \dfrac{a}{{{V}^{2}}}\,=\dfrac{a}{{{\left( {{m}^{3}} \right)}^{2}}}=N{{m}^{-2}} $$
$$ a=N{{m}^{4}}=\,\left[ M{{L}^{1}}{{T}^{-2}} \right]\left[ {{M}^{o}}{{L}^{4}}{{T}^{o}} \right]=\left[ M{{L}^{5}}{{T}^{-2}} \right] $$
Hence, unit of $$a$$ is $$\left[ M{{L}^{5}}{{T}^{-2}} \right]$$
Given that,
$$\dfrac{\Delta v}{v}\times 100$$ is the % error in velocity = 50%
Kinetic energy $$K.E=\dfrac{1}{2}m{{v}^{2}}$$
Error in the kinetic energy
$$\dfrac{\Delta K.E}{K.E}\times 100=m\times 2\dfrac{\Delta v}{v}\times 100$$
$$m$$ is as a constant
Now, percentage error
$$ \dfrac{\Delta K.E}{K.E}\times 100=2\dfrac{\Delta v}{v}\times 100 $$
$$ \dfrac{\Delta K.E}{K.E}\times 100=2\times 50$$%
$$ \dfrac{\Delta K.E}{K.E}\times 100=100$$%
Hence, the error in the measurement of kinetic energy is $$100$$%
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