Question

A rod has uniformly distributed charge $24\,\mu$C and length $10$ cm. Find force on $1\,\mu$C particle placed at a distance $2$ cm from one end of the rod.

• A. $70$ N
• B. $10.5$ N
• C. $90$ N
• D. $25$ N

Hint:

For a finite charged rod, always use integration or the standard force formula instead of Coulomb’s law for point charges.

Answer 1

The Correct Option is C

Step 1: Linear charge density of the rod.
\[ \lambda = \dfrac{Q}{L} = \dfrac{24 \times 10^{-6}}{0.1} \,\text{C/m} \]
Step 2: Expression for electric force due to a uniformly charged rod.
Force on charge $q$ placed at distance $x$ from one end of rod is:
\[ F = k q \lambda \left(\dfrac{1}{x} - \dfrac{1}{x + L}\right) \]
Step 3: Substituting given values.
\[ q = 1 \times 10^{-6}\,\text{C}, \quad x = 2 \times 10^{-2}\,\text{m}, \quad L = 10 \times 10^{-2}\,\text{m} \]
\[ F = 9 \times 10^9 \times 10^{-6} \times \dfrac{24 \times 10^{-6}}{0.1} \left(\dfrac{1}{2 \times 10^{-2}} - \dfrac{1}{12 \times 10^{-2}}\right) \]
Step 4: Final calculation.
\[ F = 9 \times 24 \times \dfrac{5}{12} = 90\,\text{N} \]