A) 0.02
B) 0.3
C) 0.2
D) 1.3
Correct Answer: C
Solution :
Potential gradient \[x=\frac{e}{(R+{{R}_{h}}+r)}.\frac{R}{L}\] \[=\frac{3}{(20+10+0)}\times \frac{20}{10}=0.2\]You need to login to perform this action.
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