Electrostatics Question 307
Question: The electric intensity due to a dipole of length 10 cm and having a charge of $ 500\mu C $ , at a point on the axis at a distance 20 cm from one of the charges in air, is [CBSE PMT 2001]
Options:
A) $ 6.25\times 10^{7} $ N/C
B) $ 9.28\times 10^{7} $ N/C
C) $ 13.1\times 11^{11} $ N/C
D) $ 20.5\times 10^{7} $ N/C
Show Answer
Answer:
Correct Answer: A
Solution:
By using $ E=9\times 10^{9}.\frac{2pr}{{{(r^{2}-l^{2})}^{2}}}; $
where $ p=(\text{5}00\times \text{1}{{0}^{\text{6}}})\times (\text{1}0\times \text{1}{{0}^{\text{2}}})=\text{5}\times \text{1}{{0}^{\text{5}}} $ ,
r = 25 cm = 0.25 m, l = 5 cm = 0.05 m $ E=\frac{9\times 10^{9}\times 2\times 5\times {{10}^{-5}}\times 0.25}{{{{{{(0.25)}^{2}}-{{(0.05)}^{2}}}}^{2}}} $
$ =6.25\times 10^{7}N/C $
