Electrostatic potential

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512 identical drops of mercury are charged to a potential of 2V each. The drops are joined to form a single drop. The potential of this drop is______ V.

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Solution

$\begin{array}{l} C o n s e r v i n g v o l u m e, \\ V_{big} = 512 V_{small} \\ \Rightarrow \frac{4}{3} π R^{3} = 8^{3} \frac{4}{3} π r^{3} \\ \Rightarrow R = 8 r \\ N o w, f o r p o t e n t i a l, \\ V_{small} = \frac{K q}{r} a n d \\ V_{big} = \frac{K q^{'}}{R} \\ h e r e c o n s e r v i n g c h a r g e g i v e s, q^{'} = 512 q \\ \Rightarrow V_{big} = \frac{K \times 512 q}{8 r} = \frac{512}{8} \frac{Kq}{r} = 64 V_{small} = 64 \times 2 \\ \Rightarrow V_{big} = 128 volt \end{array}$

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Electrostatic potential

Remember concepts with our Masterclasses.

512 identical drops of mercury are charged to a potential of 2V each. The drops are joined to form a single drop. The potential of this drop is______ V.

Conserving volume, Vbig =512Vsmall ⇒43πR3=8343πr3⇒R=8rNow, for potential,Vsmall =Kqr andVbig =Kq'Rhere conserving charge gives ,q'=512q⇒Vbig =K×512q8r=5128Kqr=64Vsmall =64×2⇒Vbig =128 volt

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