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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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Four resistors , R1=12ohms, R2= 15 ohms , R3= 18 ohms and R4=10 ohms are connected to 6 v battery in series what is the total re
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3 years ago
HELP ME PLS
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Answer

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3 0
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Why does the rider continue to go the same height on the ramp every time and would continue to do that forever?
Aleks [24]
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3 0
3 years ago
Dave wishes to get produce from the store, which is a distance D- 2.2 km east of his house. Dave drives his bike to the store at
Neporo4naja [7]

Answer:

a) t_1=338.4615\ s

b) t=1386.0805\ s

c) t=23.1013\ min

d) d=4400\ m

e) Since Dave starts from house and finally returns to the house so displacement is zero.

Explanation:

Given:

  • distance between the house and the store, s=2.2\ km=2200\ m
  • speed of driving from house to store, v_1=-6.5\ m.s^{-1}
  • speed of driving back from store to house, v_2=2.1\ m.s^{-1}

Since the store is located towards east from his house and the velocity in this direction is taken negative and contrary to this the velocity in the west direction is taken positive.

a)

time taken in reaching the store:

t_1=\frac{s}{v_1}

t_1=\frac{-2200}{-6.5}

t_1=338.4615\ s

b)

Now the time taken in returning form the store:

t_2=\frac{s}{v_2}

t_2=\frac{2200}{2.1}

t_2=1047.6190\ s

therefore total time taken by the trip:

t=t_1+t_2

t=338.4615+1047.6190

t=1386.0805\ s

c)

the time taken by the trip in minutes:

t=\frac{1386.0805}{60}

t=23.1013\ min

d)

distance travelled in the whole trip:

d=2\times s

d=2\times 2200

d=4400\ m

e)

Since Dave starts from house and finally returns to the house so displacement is zero.

3 0
3 years ago
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