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elena-14-01-66 [18.8K]

2 years ago

9

A positive test charge q is released from rest at distance r away from a charge of +Q and a distance 2r away from a charge of +2

Q.
How will the test charge move immediately after being released?

A. to the left
B. to the right
C. stay still
D. other

2 answers:

svlad2 [7]2 years ago

7 0

Answer:

The test charge will move to the left side after being released.

Explanation:

Given that,

Test charge = q

First charge = Q

Second charge = 2Q

Distance of point charge from first charge = r

Distance of point charge from second charge = 2r

Let the charge Q placed at left side and charge 2Q placed at right side.

We need to calculate the force due to first charge

Using formula of force

$F=\dfrac{kq_{1}q}{r^2}$

Put the value into the formula

$F=\dfrac{kQq}{r^2}$ ...(I)

We need to calculate the force due to second charge

Using formula of force

$F'=\dfrac{k2Qq}{4r^2}$ ...(I)

Divided equation (I) by equation (II)

$\dfrac{F}{F'}=\dfrac{\dfrac{kQq}{r^2}}{\dfrac{k2Qq}{4r^2}}$

$\dfrac{F}{F'}=\dfrac{1}{2}$

$F=\dfrac{1}{2}F'$

Hence,The test charge will move to the left side after being released.

Svetradugi [14.3K]2 years ago

3 0

Answer:

B. to the right

Explanation:

Given:

test charge +q

distance of the test charge from +Q, r

distance of test charge from +2Q, 2r

<u>Force on the test charge due to +Q:</u>

$F_1=k.\frac{Q.q}{r^2}$

<u>Force on the test charge due to +Q:</u>

$F_2=k.\frac{2Q.q}{(2r)^2}$

$F_2=k.\frac{Q.q}{2r^2}$

Since all the charges are positive here, so they will try to repel the test charge away. And the force due to charge +Q will be greater so initially the test charge will move rightwards away from the +Q charge.

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