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Helen [10]
3 years ago
7

When two point charges are 2.0 cm apart, each one experiences a 1.0-N electric force due to the

Physics
2 answers:
Nitella [24]3 years ago
6 0

Answer:

option (E)

Explanation:

F = 1 N

r = 2 cm

r' = 8 cm

F' = ?

According to the Coulomb's law

F \alpha \frac{1}{r^{2}}

So, \frac{F'}{F}=\left (\frac{r}{r'}  \right )^{2}

\frac{F'}{1}=\left (\frac{2}{8}  \right )^{2}

F' = 0.063 N

Stella [2.4K]3 years ago
4 0

Answer:

0.063N

Explanation:

According to coloumbs law, the force of attraction F between two charges is directly proportional to the product of the charges q and Q and inversely proportional to the square of the distance 'r' between them.

F = kqQ/r²

where k is the coulombs constant

If two point charges are 2.0 cm apart, each one experiences a 1.0-N electric force due to the other charge, the equation becomes

1.0 = kqQ/0.02²

kqQ = 0.0004... (1)

If they are moved to a new separation of 8.0 cm, the electric force on each of them will be

F = kqQ/0.08²

0.0064F = kqQ

kqQ = 0.0064F ...(2)

Dividing equation 2 by 1 gives;

kqQ/kqQ = 0.0064F/0.0004

1 = 16F

F = 1/16

F = 0.0625N

The electric force on each of them

is closest to 0.063N

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