Explanation: Hints: তড়িৎ ক্ষেত্রের প্রাবলা, \( E = \frac{1}{4\pi\epsilon_0} \times \frac{q}{r^2} \) Solve: অসীম সংখ্যক চার্জের তড়িৎ প্রাবলা, \( E = E_1 + E_2 + \ldots + E_\infty \) \( = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r_1^2} + \frac{1}{4\pi\epsilon_0} \frac{q_2}{r_2^2} + \ldots + \frac{1}{4\pi\epsilon_0} \frac{q_\infty}{r_\infty^2} \) \( = \frac{1}{4\pi\epsilon_0} \left( \frac{1}{1^2} + \frac{1}{2^2} + \ldots + 0 \right) \) [যেহেতু \( q_1 = q_2 = \ldots = q_\infty = 1C \)] \( = \frac{1}{4\pi\epsilon_0} \sum \frac{1}{n^2} = \frac{1}{4\pi\epsilon_0} \times \frac{\pi^2}{6} = 1.48 \times 10^{10} \) Ans. (E)