A) 1/R
B) \[1/{{R}^{2}}\]
C) R
D) \[{{R}^{2}}\]
Correct Answer: A
Solution :
We knows \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2\lambda }{R}\] Hence \[E\propto \frac{1}{R}\]
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