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nydimaria [60]
3 years ago
10

A sphere with a charge q is fixed at the bottom left corner of the right triangle shown in the figure. Points P and R are at the

location shown. A test charge is place at point P, and the charge-sphere system has potential energy U0. If the test charge is then moved to point R, what will the potential energy of the charge-sphere system be? ​

Physics
1 answer:
Alexus [3.1K]3 years ago
5 0

Answer:

the final potential energy of this system is 3U0/10

Explanation:

We are given

charge at left end  and another test charge at point p

Potential energy is given by  = \frac{k*Q1*Q2 }{R}

where k is electrostatics constant = 9 *10^9

Q1 = first charge , Q2=  test charge

R= distance between charges

potential at point p

U0 = k*Q1*Q2 /3 ⇒ kq1q2 = 3U0 ..............1

now the test charge moves to point R

using Pytahgoreou theorem

R(distance) = \sqrt{8^2 + 6^2} = 10

New Potential energy

U1 = kq1*q2 / 10

substituting  kq1q2 = 3U0 from 1

U1 = 3U0/10

So this is the final potential energy of this system.

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Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

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Explanation:

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To understand it better, regarding to the equivalence principle, Einstein formulated the following:  

A gravitational force and an acceleration in the opposite direction are equivalent, both have indistinguishable effects.  Because the laws of physics must be accomplished in all frames of reference.

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