Which of the following are permissible amounts of charge for an object to have?

Question

Which of the following are permissible amounts of charge for an object to have? Select all that are correct (and none that are incorrect) to receive credit.
(You should assume the elementary charge value is exactly e=1.6·10-19 C, and that the values below have no uncertainty.)

-1/2 C

1.6·e

1.4·10-19 C

4·e

1.4657563 nC

-1.1 C

SELECT ANY AND ALL THAT APPLY

Two point charges lie on the x axis. A charge of -6.5 C is at the origin, and a charge of + 7.5 C is at x = +10.0 cm.
(In this 1D problem with a clearly defined axis, you may indicate the direction of the electric field with + = to the right, - = to the left.)

(a) What is the net electric field (size & dir) at x = -2.0 cm?
N/C
(b) What is the net electric field (size & dir) at x = +2.0 cm?
N/C

Explanation / Answer

(a).The required charge is integral multiples of charge of electron .

Answer is 1.6

(b).Charge q = -6.5 x10 -6 C

                 q ' = 7.5 x10 -6 C

Separation of the two charges q and q ' is r = 10 cm = 0.1 m

Coulomb's constant K = 8.99 x10 9 Nm 2/C 2

(a). Electric field at x = -2 cm due to charge q is E = Kq/x 2

Where x = 2 cm = 0.02 m

Substitute values you get E =(8.99x10 9)(6.5 x10 -6 ) / 0.02 2

                                            = 1.46 x10 8 N/C

Electric field at x = -2 cm due to charge q ' is E ' = Kq/(r+x) 2

Where x = 2 cm = 0.02 m

Substitute values you get E ' =(8.99x10 9)(7.5 x10 -6 ) / (0.1+0.02) 2

                                            = 4.68 x10 6 N/C

E and E ' are opposite to each other.

So, the net electric field at x = -2cm is = E -E '

                                                            = 141.32 x10 6 N/C    It is along positive x direction.

(b). Electric field at x = -2 cm due to charge q is E = Kq/x 2

Where x = 2 cm = 0.02 m

Substitute values you get E =(8.99x10 9)(6.5 x10 -6 ) / 0.02 2

                                            = 1.46 x10 8 N/C

Electric field at x = -2 cm due to charge q ' is E ' = Kq/(r+x) 2

Where x = 2 cm = 0.02 m

Substitute values you get E ' =(8.99x10 9)(7.5 x10 -6 ) / (0.1-0.02) 2

                                            = 10.53 x10 6 N/C

E and E ' are in same direction.

So, the net electric field at x = 2cm is = E +E '

                                                            = 156.53 x10 6 N/C    It is along negative x direction.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.