A spherical conductor of radius 10 cm has a charge of 3.2 × 10⁻⁷ C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere ?

\(\left(\frac{1}{4 \pi \epsilon_{0}} = 9 \times 10^{9} \text{Nm}^{2}/\text{C}^{2}\right) \)

Question

Question

This question was previously asked in

NEET 2020 Official Paper (Held On: 13 September, 2020)

Answer 1

Explanation:

The electric field outside a conducting sphere

\(E = \frac{1}{{4\pi { \in _0}}}\frac{Q}{{{r^2}}} \)

here we have Q as the charge and r as the distance.

Given: Q = 3.2 × 10−7 C

r = 15 cm

The electric field is written as;

\(E = \frac{1}{{4\pi { \in _0}}}\frac{Q}{{{r^2}}}\)

⇒ \( E= \frac{{9 × {{10}^9} × 3.2 × {{10}^{ - 7}}}}{{225 × {{10}^{ - 4}}}}\)

⇒ E = 0.128 × 106

⇒ E = 1.28 × 105 N/C

Hence, option 4) is the correct answer.