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Pani-rosa [81]
3 years ago
14

Suppose the electric field in problems 2 was caused by a point charge. The test charge is moved to a distance twice as far from

the charge. What is the magnitude of the force that the field exerts on the test charge now ?
Physics
1 answer:
mojhsa [17]3 years ago
8 0

Answer:

it is reduced four times.

Explanation:

By definition, the electric field is the force per unit charge created by a charge distribution.

If the charge creating the field is a point charge, the force exerted by it on a test charge, must obey Coulomb´s Law, so, it must be inversely proportional to the square of the distance between the charges.

So, if the distance increases twice, as the force is inversely proportional to the square of the distance, and the square of 2 is 4, this means that the magnitude of the force exerted on the test charge must be 4 times smaller.

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Answer:

926 N

Explanation:

Metric unit conversion:

R = 18 cm = 0.18 m

r = 5 cm = 0.05 m

The pressure exerted by the F = 12000N car on the wider arm would be ratio of the gravity over area

P = F/A = \frac{F}\pi R^2} = \frac{12000}{2*\pi*0.18^2} = 117892 Pa

The pressure must be the same on the smaller pressure for it to be able to start lifting the car. We can calculate the force f acting on it:

f = Pa = P\pi r^2 = 117892 * \pi * 0.05^2 = 926 N

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A 19.0-kg cart is moving with a velocity of 7.20 m/s down a level hallway. A constant force of -13.0 N acts on the cart and its
gulaghasi [49]

Answer:

a. -369.36J

b. -123.9J

c. 9.52m

Explanation:

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