The electirc field intensity at a distance of 20 cm from the centre of a small conducting sphere is $1.5 × 10^3\, NC^{-1}$ and is directed towards the sphere. The charge on the sphere is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → - 6.6 nC
Given:
$E = 1.5 \times 10^{3} \, N/C$
$r = 20 \, cm = 0.20 \, m$
$\varepsilon_0 = 8.85 \times 10^{-12} \, C^2 N^{-1} m^{-2}$
Formula:
$E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}$
Rearranging:
$Q = E \cdot 4 \pi \varepsilon_0 r^2$
Substitute:
$Q = (1.5 \times 10^{3})(4 \pi \times 8.85 \times 10^{-12})(0.20)^2$
$Q = 6.67 \times 10^{-9} \, C$
Answer: $Q = 6.67 \times 10^{-9} \, C$