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lubasha [3.4K]
3 years ago
14

A small object has charge Q. Charge q is removed from it and placed on a second small object The two objects are placed I m apar

t. For the force that each object exerts on the other to be a maximum, what should q be? a) q 20 c) q Q/2 d) q /4
Physics
1 answer:
Whitepunk [10]3 years ago
4 0

Answer:

q = Q/2

option c is correct

Explanation:

given data

charge = Q

distance r = 1 m

charge = q

to find out

what should q be

solution

we have given charge q is remove so

q1 = Q-q

q2 = q

we know by coulombs law force

force = kq1q2 / r²   .............1

put here value and we know k is constant

F = k (Q-q) q / 1

now take derivative charge q and put = 0

dF/dq = k d(Q-q)q / dq

so  k (Q-q)q =0

Q = 2q

q = Q/2

so option c is correct

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How do you work out the potential diffrance
timurjin [86]

Answer:

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two ships leave a port at the same time. The first ship sails on a bearing of 40 degrees at 18 knots and the second at a bearing
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Answer:

The distance between the ships is 87.84 km.

Explanation:

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v=\sqrt{324+676}

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Put the value into the formula

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7 0
3 years ago
I''l give brainliest. Please help. I hold 2 objects about 0.1 meters apart. What is the electrostatic force between the two obje
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Answer:

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2 years ago
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Answer: 0.04139m

Explanation:

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Then, we calculate the force which will be:

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x = mg/k

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x = 0.04139m.

The spring stretched for 0.04139m.

4 0
3 years ago
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