Question

Find electric field intensity \( \vec{E} \) at the centre of the circle shown in the figure.

• A. \( \dfrac{KQ}{R^2}\hat{i} + \dfrac{KQ}{R^2}\hat{j} \)
• B. \( -\dfrac{\sqrt{3}KQ}{R^2}\hat{i} + \dfrac{KQ}{R^2}\hat{j} \)
• C. \( \dfrac{KQ}{R^2}\hat{i} + \dfrac{\sqrt{3}KQ}{R^2}\hat{j} \)
• D. \( \dfrac{\sqrt{3}KQ}{R^2}\hat{i} + \dfrac{\sqrt{3}KQ}{R^2}\hat{j} \)

Hint:

For symmetric charge distributions:
Always resolve fields into components
Use symmetry to cancel components wherever possible

Answer 1

The Correct Option is B

Step 1: Equate Rate Constants. \[ 10^4 e^{-24000/T} = 10^6 e^{-30000/T} \]
Step 2: Simplify. Divide by \(10^4\) and exponential terms: \[ \frac{e^{-24000/T}}{e^{-30000/T}} = \frac{10^6}{10^4} = 100 \] \[ e^{(-24000 + 30000)/T} = 100 \] \[ e^{6000/T} = 100 \]
Step 3: Solve for T. Take natural log (ln): \[ \frac{6000}{T} = \ln(100) = 2 \ln(10) \] Using \(\ln(10) \approx 2.303\): \[ \frac{6000}{T} = 2(2.303) = 4.606 \] \[ T = \frac{6000}{4.606} \approx 1302.6\,\text{K} \] Rounding to nearest integer: 1303 K.