Answer:
Step-by-step explanation:
capacitance of the capacitor C = 4πε₀R
= 1 x .103 / (9 x 10⁹ ) [ 1/ 4πε₀ = 9 x 10⁹ . ]
= .0114 x 10⁻⁹ F
potential V = 4.5 x 10³ v
charge Q = CV
.0114 x 10⁻⁹ x 4.5 x 10³
= .0515 x 10⁻⁶ C
charge on one electron = 1.6 x 10⁻¹⁹
no of electrons to be removed
= .0515 x 10⁻⁶ / 1.6 x 10⁻¹⁹
= .032 x 10¹³
3.2 x 10¹¹ electrons.