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Alex_Xolod [135]

4 years ago

7

"A positively-charged particle is held at point A between two parallel metal plates. The plate on the left has a net positive ch

arge q and the plate on the right has a net negative charge -q. The particle is then moved to point B. How does the electric potential energy at point A compare with that at point B?

1 answer:

defon4 years ago

8 0

Answer:

Explanation

A positively charged particle produces a positive electric potential while a negatively

charged particle produces a negative electric potential.

Given two parallel plates one is positive and the other is negative and a positively charge particle is place at point A.

The electric potential V at point P is the algebraic sum of the electric potentials contributed by the two parallel plates charges

V=Kq1/r+ Kq2/r

Since q1 is positive and q2 is negative and they have the same magnitude q.

Then, at position A,

Let assume the total distance between the parallel plate is x

Then, the positive plate should be at distance 'a' from, the other is at x-a.

V= Kq/a-kq/(x-a)

V=kq(1/a-1/(x-a))

V=kq(x-a-a/a(x-a))

V=kq(x-2a)/(ax-a^2)

Then, at position B

Let assume the total distance between the parallel plate is x

Then, the positive plate should be at distance 'b' from, the other is at x-b.

V= Kq/b-kq/(x-b)

V=kq(1/b-1/(x-b))

V=kq(x-b-b/b(x-b))

V=kq(x-2b)/(bx-b^2)

Close to the positive plate, the

potential has very large positive values. Close to the negative charge, the potential has very large negative values.

They have different potential.

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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

Gnesinka [82]

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 $U_f -U_i = 16 *10^{-7} J$

Explanation:

From the question we are told that

The radius of the soap bubble is $R_o = 7.45 \ cm = \frac{7.45}{100} = 0.0745 \ m$

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

$U_i = \frac{V_1^2 R_o }{2k }$

The final electric potential is mathematically represented as

$U_f = \frac{V_2^2 R_1 }{2k }$

The initial potential is mathematically represented as

$V_1 = \frac{kQ}{R_o}$

The final potential is mathematically represented as

$V_2 = \frac{kQ}{R_1}$

Now

$\frac{V_2}{V_1} = \frac{R_o}{R_1}$

substituting values

$\frac{V_2}{V_1} = \frac{7.45}{1.4155} = \frac{1}{0.19}$

=> $V_2 = \frac{V_1}{0.19}$

So

$U_f = \frac{V_1^2 R_2 }{2k * 0.19^2}$

Therefore

$U_f -U_i = \frac{V_1^2 R_2 }{2k * 0.19^2} - \frac{V_1^2 R_o }{2k }$

$U_f -U_i = \frac{V_1^2}{2k} [\frac{ R_1 }{ * 0.19^2} - R_o]$

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

substituting values

$U_f -U_i = \frac{307^2}{9 * 10^{9}} [\frac{ 0.014155 }{ 0.19^2} - 0.0745]$

$U_f -U_i = 16 *10^{-7} J$

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Answer:

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Answer:

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