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Ierofanga [76]
2 years ago
8

Two charges are separted by a distance d and exert mutual attractive forces of f on each other. If the charges are separated by

a distate of 2d what are the new mutual forces?
Physics
1 answer:
ASHA 777 [7]2 years ago
3 0

The new force between the charges when the distance become 2d will be  F'=4F

<h3>What is electrostatic force?</h3>

When two charged particles are separated by the distance d then the force of attraction or repultion acts on the charged particle depending upon the nature of the charge whether it is positive or negative.

The formula for the electrostatic force is given by

F=\dfrac{kq_1q_2}{d^2}

Now if the value of d becomes 2d then the formula will become:

F'=\dfrac{kq_1q_2}{(2d)^2}=\dfrac{kq_1q_2}{4d}

\dfrac{F'}{F}=\dfrac{1}{4}

F=4F'

Hence the new force between the charges when the distance become 2d will be  F'=4F

To know more about electrostatic force follow

brainly.com/question/17692887

#SPJ4

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\Delta \vec{p} = 0\\\\m\vec{v}_i = m\vec{v}_f\\m_S v_S\hat{i} + m_A v_A\hat{j} = m_S \times 7 \cos{(40)} \hat{i} + m_S \times 7 \sin{(40)} \hat{j} + m_A \times 9.4 \cos{(18)} \hat{i} - m_A \times 9.4 \sin{(18)} \hat{j}.

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Their kinetic energy changed by

K_f - K_i = \left( \frac{1}{2}m_s v_{fs}^2 + \frac{1}{2}m_a v_{fa}^2 \right) - \left( \frac{1}{2}m_s v_{is}^2 + \frac{1}{2}m_a v_{ia}^2 \right) =  \left( \frac{1}{2}\times 73 \times 7^2 + \frac{1}{2}\times 57 \times 9.4^2 \right) - \left( \frac{1}{2}\times 73 \times 12.34^2  + \frac{1}{2}\times 57 \times 2.86^2 \right) \approx \mathbf{-1484.418 J}.

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