213k views
1 vote
n isolated charged soap bubble of radius R0=7.45 cmR0=7.45 cm is at a potential of V0=307.0 volts.V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r

User Ahmetunal
by
3.9k points

1 Answer

3 votes

Complete Question

An isolated charged soap bubble of radius R0 = 7.45 cm is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r

Answer:

The difference is
U_f -U_i = 16 *10^(-7) J

Step-by-step explanation:

From the question we are told that

The radius of the soap bubble is
R_o = 7.45 \ cm = (7.45)/(100) = 0.0745 \ m

The potential of the soap bubble is
V_1 =307.0 V

The new radius of the soap bubble is
R_1 = 0.19 * 7.45=1.4155\ cm = 0.014155 \ m

The initial electric potential is mathematically represented as


U_i = (V_1^2 R_o )/(2k )

The final electric potential is mathematically represented as


U_f = (V_2^2 R_1 )/(2k )

The initial potential is mathematically represented as


V_1 = (kQ)/(R_o)

The final potential is mathematically represented as


V_2 = (kQ)/(R_1)

Now


(V_2)/(V_1) = (R_o)/(R_1)

substituting values


(V_2)/(V_1) = (7.45)/(1.4155) = (1)/(0.19)

=>
V_2 = (V_1)/(0.19)

So


U_f = (V_1^2 R_2 )/(2k * 0.19^2)

Therefore


U_f -U_i = (V_1^2 R_2 )/(2k * 0.19^2) - (V_1^2 R_o )/(2k )


U_f -U_i = (V_1^2)/(2k) [( R_1 )/( * 0.19^2) - R_o]

where k is the coulomb's constant with value
9*10^(9) \ kg\cdot m^3\cdot s^(-4)\cdot A^2.

substituting values


U_f -U_i = (307^2)/(9 * 10^(9)) [( 0.014155 )/( 0.19^2) - 0.0745]


U_f -U_i = 16 *10^(-7) J

User Roland Jansen
by
3.7k points