Question

A spherical conductor of radius 10 cm has a charge of \(3.2 \times 10^{-7} C\) distributed uniformly. What is the magnitude of electric field at a point \(15\,cm\) from the centre of the sphere?
\((\frac {1}{4\pi \epsilon_o} = 9 \times 10^9 Nm^2/C^2)\)

• A. $1.28 \times 10^4\,N/C$
• B. $1.28 \times 10^5\,N/C$
• C. $1.28 \times 10^6\,N/C$
• D. $1.28 \times 10^7\,N/C$

Answer 1

The Correct Option is B

To find the magnitude of the electric field at a point 15 cm from the center of a spherical conductor with radius 10 cm, we apply the concept of the electric field due to a point charge. According to Gauss's law, the electric field outside a uniformly charged spherical conductor can be treated as if all the charge is concentrated at its center. Given:

• Charge, \(q = 3.2 \times 10^{-7} \, C\)
• Distance from the center, \(r = 15 \, cm = 0.15 \, m\)
• Coulomb's constant, \(k = \frac{1}{4\pi \epsilon_o} = 9 \times 10^9 \, Nm^2/C^2\)

The formula for the electric field \(E\) at a distance \(r\) from a point charge is:

\(E = \frac{k \cdot q}{r^2}\)

Substituting the given values:

\(E = \frac{9 \times 10^9 \times 3.2 \times 10^{-7}}{(0.15)^2}\)

\(E = \frac{9 \times 3.2 \times 10^2}{0.0225} \, N/C\)

\(E = \frac{28.8 \times 10^2}{0.0225} \, N/C\)

\(E = 1.28 \times 10^5 \, N/C\)

Thus, the magnitude of the electric field at a point 15 cm from the center of the spherical conductor is \(1.28 \times 10^5 \, N/C\).

The correct answer is: \(1.28 \times 10^5 \, N/C\).