Question

A spherical conductor of diameter \(3\) mm is kept in a uniform electric field of intensity \(4\times 10^{7}\) N/C. The maximum charge on the conductor is \(\left(\frac{1}{4\pi \epsilon _{0}}=9\times 10^{9}SIunits\right)\)

• A. \(2\times 10^{-8}C\)
• B. \(10^{-8}C\)
• C. \(4\times 10^{-8}C\)
• D. \(3\times 10^{-8}C\)

Hint:

The surface field of a charged sphere is kQ over r squared; the maximum charge corresponds to the given field at the surface.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The field at the surface of a spherical conductor is \(E=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}\).

Step 2: Values
Diameter \(3\) mm gives \(r=1.5\times10^{-3}\) m.
\(E=4\times10^{7}\) N/C, \(k=9\times10^{9}\).

Step 3: Calculation
\(Q=\frac{Er^{2}}{k}=\frac{4\times10^{7}\times(1.5\times10^{-3})^{2}}{9\times10^{9}}\)
\(=\frac{4\times10^{7}\times2.25\times10^{-6}}{9\times10^{9}}=10^{-8}\) C

Step 4: Compare the options
This is option (B).

Final Answer:
The maximum charge is 10 to the power minus 8 coulomb. \[ \boxed{\text{(B) }10^{-8}\,\text{C}} \]