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A spherical drop of water carrying a charge of 30 pChas a potential of 500 Vat its surface (with V = 0at 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?

Short Answer

Expert verified
  1. The radius of the drop is 0.54 mm.
  2. The potential at the surface of the new drop is 790 V.

Step by step solution

01

Given data:

The Coulomb鈥檚 constant,k=9109N.m2/C2

The charge, q=30pC=3010-12C

The potential, V = 500 V

02

Understanding the concept:

You have to use the concept of the electric potential to calculate the radius of the drop. The expression for the electric potential on the surface of the spherical drop is given by,

V=kqR 鈥.. (1)

Where, k is the Coulomb鈥檚 constant, q is the charge on the spherical drop, and R is the radius of the spherical drop.

03

(a) The radius of the drop:

The radius of the drop is defined by rearranging equation (1).

R=kqV

Substitute known values in the above equation.

R=9109Nm2/C23010-12C500V=0.5410-3m=0.54mm

Hence, the radius of the spherical drop is 0.54 mm.

04

(b) The potential at the surface of the new drop:

Calculatethe potential at the surface of the new drop if two such drops of the same charge and radius combine to form a single spherical drop.

The volume of the new drop is given by,

V ' = 2V

Here, V is the volume of the individual spherical drops.

The volume of the spherical drop is given by,

V=43蟺搁3

Here, R is the radius of the spherical drop.

The volume of the new drop is given by,

V'=43蟺搁'3

Here, R' is the radius of the new drop.

Substitute 43蟺搁'3 for V ' and 43蟺搁3 for V in equation V ' = 2V as follows:

43蟺搁'3=243蟺搁3R'3=2R3R'=213R

Substitute 0.54 mm for R in above equation as follows:

R'=2130.54mm10-3m1mm=6.810-4m

The expression for the electric potential is given by,

V'=k2qR'=9109Nm2/C223010-12C6.810-4m=794V

After rounding off to two significant figures the electric potential is 790 V.

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