Q 12-01-022JEE MainMedium
A small ball of mass $2$ g and charge $+5\ \mu$C hangs from a thread in a uniform horizontal electric field. In equilibrium the thread makes $45°$ with the vertical. Find the field strength in N/C. ($g = 10\ \text{m/s}^2$)
Numerical value type. Enter your answer.
Answer: 4000
Concept. In equilibrium $\tan\theta = \dfrac{qE}{mg}$.
Step 1. Solve. $\tan45° = 1$:
$$E = \frac{mg}{q} = \frac{2 \times 10^{-3} \times 10}{5 \times 10^{-6}} = 4000\ \text{N/C}$$
Answer. $4000$.
Solution by Sreeraj P, M.Sc Physics