\(E=\frac{F}{q} \text { and } F=Q \cdot V \cdot B\)
\(\Rightarrow E=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} \times B-(1)\left(\text { as } V=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\right)\)
Also, \(E=\varepsilon_0 E^2=\frac{1}{2} \varepsilon_0 E^2+\frac{1}{2} \varepsilon_0 \times \frac{B^2}{\varepsilon_0 \mu}\).
\(\Rightarrow E=\frac{1}{2} \varepsilon_0 E^2+\frac{1}{2} B^2 / \mu_0\)
\(\therefore\) It depends on both Electric and Magnetic field.
(આપેલ :શુન્યાવાકાશની પરમીટીવીટી $\varepsilon_{0}=8.85 \times 10^{-12} \,C ^{2} N ^{-1}- m ^{-2}$, શુન્યાવાકાશમાં પ્રકાશની ઝડપ $\left.c=3 \times 10^{8} \,ms ^{-1}\right)$