Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they are separated by 13 cm. The Coulomb constant is 8.99 × 109 N · m2 /C 2 . Answer in units of N. 032 (part 2 of 2) 10.0 points If the two charges have equal magnitude, what is the magnitude of each charge for the original force of 15 N? Answer in units of C.

[tex]F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N[/tex]

Part (b)

the magnitude of each charge, if they have equal magnitude

[tex]F = \frac{KQ^2}{r^2}[/tex]

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

[tex]F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q = \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C[/tex]

andromache andromache · Answer 2 · 2022-03-25T09:27:12+01:00

(a) the force is 8.876 N.

(b) the magnitude of each charge is 4.085 μC.

Coulomb's law:

Since

coulomb's constant, K = [tex]8.99 \times 10^9 N.m^2/C^2[/tex]

distance between two charges, r = 10 cm = 0.1 m

the force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

We know that

[tex]F = Kq_1q_2/r^2[/tex]

Here we assume Kq1q2 = C

So,

[tex]F = C/r^2\\\\C = Fr^2\\\\F_1r_1^2 = G_2r_2^2\\\\F_2 = F_1r_1^2\div r_2^2\\\\= 15\times 0.1^2\div 0.13^2[/tex]

= 8.876 N

b.

Now

We know that

[tex]F = KQ^2\div r^2[/tex]

here,

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

[tex]F = KQ^2\div r^2\\\\Q = \sqrt Fr^2\div k\\\\= \sqrt 15\times (0.1)^2 \div 8.99\times 10^9\\\\= 4.085\times 10^{-6}C[/tex]

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