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2 years ago

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Two charged particles, with charges q1 = q and q2 = 4q are located at a distance d = 2cm apart on the x axis A third charged par

ticle, with charge q3 = q is placed on the x axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3. Find the position of charge 3 when q = 10n C

1 answer:

weqwewe [10]2 years ago

3 0

Answer:

0.667 cm

Explanation:

|F_1on3| = |F_2on3|

k×q1×q3/x^2 = k×q2×q3/(d-x)^2

q1/x^2 = q2/(d-x)^2

(d-x)^2/x^2 = q2/q1

(d-x)/x = sqrt(q2/q1)

(2 - x)/x = sqrt(4×q/q)

(2 - x)/x = 2

2 - x = 2x

2 = 3x

x = 2/3

x= 0.667 cm

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a) At the equator, both the <u>centripetal force</u> $F_{c}$ and the <u>gravitational force</u> $F_{g}$ (also called true weight) are directed "downward", while the <u>normal force</u> $n_{equator}$ (also called apparent weight) is directed "upward". Therefore we have the following equation:

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Where:

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$F_{c}=m a_{c}$ being $a_{c}=0.0337 m/s^{2}$ the centripetal acceleration at the equator

According to this (1) is rewritten as:

$n_{equator}-mg=-m a_{c}$ (2)

Isolating $n_{equator}$ :

$n_{equator}=-m a_{c} + mg$ (3)

$n_{equator}=m(-a_{c}+g)$ (4)

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$n_{poles}-F_{g}=-F_{c-poles}$ (8)

Since $F_{c-poles}=0$ (8) is rewritten as:

$n_{poles}=F_{g}$ (9)

$n_{poles}=m g$ (10)

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