Physics, 22.02.2020 01:25 apalacios3503

A positive point charge q is placed on the +y-axis at y = a, and a negative point charge –q is placed on the –y-axis at y = -a. A negative point charge –Q is located at some point on the +x-axis.
a. Draw a free-body diagram showing the forces that act on the charge –Q.
b. Find the x and y components of the net force that the two charges q and –q exert on –Q. (Your answer should involve only k, q, Q, a, and the x-coordinate of the charge -Q)
c. What is the net force on the charge –Q when it is at the origin ( x = 0 )?

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Answer from: michaelgibsonsto

a) See free-body diagram in attachment

b) $F_x=0, F_y=\frac{2akqQ}{(a^2+x^2)^{\frac{3}{2}}}$

c) $F=\frac{2kqQ}{a^2}$

Explanation:

The free-body diagram of the situation can be found in attachment.

We have:

- The charge of +q is located at y = a

- The charge of -q is located at y = -a

- The charge -Q is located at x = a

We observe that:

- The force exerted by charge +q on charge -Q is attractive, since the two charges have opposite sign, so it points to the north-west direction

- The force exerted by charge -q on charge -Q is repulsive, since the two charges have same sign, so it points to the north-east direction

So, the net force points north.

We start by evaluating the situation on the x-direction first.

We observe that:

- The two charges +q and -q have same magnitude

- Also, they are located at exactly same distance from charge -Q

This means that the x-components of the force that each charge exerts on -Q are equal, but opposite in direction: therefore, they cancel each other, so the net force on the x-direction is zero:

F_x = 0

Instead, we observe that the y-components of the force that each charge exerts on -Q are both upward, therefore they add together.

The distance between charge +q and charge -Q is:

$r=\sqrt{a^2+x^2}$

where x is the location of the charge -Q.

The force between any of the two charges q, -q and charge -Q is given by (magnitude):

$F=\frac{kqQ}{r^2}=\frac{kqQ}{a^2+x^2}$

However, we are only interested in the y-component of the force, which is given by

$F_y = F cos \theta$

where $\theta$ is the angle between F_y and . By using trigonometry,

$cos \theta=\frac{a}{r}=\frac{a}{\sqrt{a^2+x^2}}$

So, substituting into the equation for F_y , we find

$F_y = \frac{kqQ}{a^2+x^2}\cdot \frac{a}{\sqrt{a^2+x^2}}=\frac{akqQ}{(a^2+x^2)^{\frac{3}{2}}}$

However, the net force is the sum of the forces due to the 2 charges, so

$F_y=\frac{2akqQ}{(a^2+x^2)^{\frac{3}{2}}}$

Here we want to find the net force on the charge -Q when this is located at the origin, so when

x = 0

We already said that the net force on the x-direction is always zero, so

F_x=0

Instead the net force on the y-direction is given by

$F_y=\frac{2akqQ}{(a^2+x^2)^{\frac{3}{2}}}$

Therefore, by substituting x = 0, we find:

$F_y=\frac{2akqQ}{(a^2+0^2)^{\frac{3}{2}}}=\frac{2akqQ}{a^3}=\frac{2kqQ}{a^2}$

And the net force is in the upward direction, towards charge +q.

A positive point charge q is placed on the +y-axis at y = a, and a negative point charge –q is place

Answer from: Quest

the assestment will probley apele with cause and effect

Answer from: Quest

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//

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