The electric field at 20 cm from the center of a conducting sphere of radius 2 cm is $4.5 × 10^3\, NC^{-1}$. The magnitude of charge on the sphere is

$2 × 10^{-8} C$

The correct answer is Option (2) → $2 × 10^{-8} C$

Given:

Distance from center, $r = 20\ \text{cm} = 0.2\ \text{m}$

Radius of sphere, $R = 2\ \text{cm} = 0.02\ \text{m}$

Electric field, $E = 4.5 \times 10^{3}\ \text{N/C}$

Electric field outside a conducting sphere:

$E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}$

Solving for $Q$:

$Q = E \cdot 4 \pi \epsilon_0 r^2$

Substitute values ($\epsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$):

$Q = 4.5 \times 10^{3} \cdot 4 \pi \cdot 8.85 \times 10^{-12} \cdot (0.2)^2$

$Q = 4.5 \times 10^{3} \cdot 4 \pi \cdot 8.85 \times 10^{-12} \cdot 0.04$

$Q \approx 4.5 \times 10^{3} \cdot 4.44 \times 10^{-12}$

$Q \approx 1.998 \times 10^{-8}\ \text{C} \approx 2.0 \times 10^{-8}\ \text{C}$

∴ Magnitude of charge on the sphere = $2.0 \times 10^{-8}\ \text{C}$