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•15 SSM ILW A spherical drop of water carrying a charge of 30 pC has a potential of 500 V at its surface (with V= 0 at infinity). (a) What is the radius of the drop? (b) If two such drops of the same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop?

•15 SSM ILW A spherical drop of water carrying a charge of 30 pC has a potential of 500 V at its surface (with V= 0 at infinity). (a) What is the radius of the drop? (b) If two such drops of the same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop?

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Figure 24-36 Problem 14. •15 SSM ILW A spherical drop of water carrying a charge of 30 pC has a potential of 50o V at its surface (with V= 0 at infinity). (a) What is the radius of the drop? (b) If two such drops of the same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop?
Transcribed Image Text:Figure 24-36 Problem 14. •15 SSM ILW A spherical drop of water carrying a charge of 30 pC has a potential of 50o V at its surface (with V= 0 at infinity). (a) What is the radius of the drop? (b) If two such drops of the same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop?
Expert Solution
Step 1

Given,

the charge on the spherical drop is Q=30pC=30×10-12C

And the potential at its surface is V=500V

 

a)

Since the potential at infinity is zero,

So potential difference between infinity and surface is

∆V=kQRVR-V∞=kQR

so the radius of the spherical drop is

R=kQVR-V∞

putting the values,

R=9×109N.m2/C2×30×10-12C500V-0V=5.4×10-4m=0.54mm

 

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