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Question Number 52845 by peter frank last updated on 13/Jan/19

Answered by peter frank last updated on 14/Jan/19

Commented by peter frank last updated on 14/Jan/19

$$absenceofvoltage$$$$mg=6\eta \pi {rv}_1......\left(i\right)$$$$presenceofappliedvoltage$$$$qE-mg=6\pi \eta {rv}_2.....\left(ii\right)$$$$divideqniibyi$$$$\frac{qE}{mg}-1=\frac{v_2}{v_1}$$$$$$

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